Is there a generalization of factoring that can be extended to the Real numbers?

This is a question that is usually answered via Ring Theory. You can think of a ring as a set where addition, substraction, and multiplication is well defined. Here, we aren't thinking of multiplication as repeated addition or addition as repeatedly adding one, rather we just care about the formal properties of addition and multiplication. Namely we want them to obey a certain list of axioms. (Commutativity, associativity, distributivity...)

In this formalism we define primes as elements $p \in R$ such that if $p \mid ab$ then $p \mid a$ or $p \mid b$. We say that $p$ divides $a$, or $p \mid a$ if there exists an element $r \in R$ such that $a=rp$. We also require $p$ not to be a "unit", which you can think of as the same thing as requiring $p\neq1$ in the usual integers. Then, in $\mathbb Q$ or $\mathbb R$ there are no primes as every non-zero element is a unit. (You can think of it intuitively as there being no primes because you can always factorize a number $x$ as $x^2\cdot\frac{1}{x}$)

There are ways of making new factorizations, just by considering instead of all the real numbers, only some of them. For example, let $$ \mathbb Z[\sqrt2] = \{a+b\sqrt2\mid a,b \in \mathbb Z\} \subseteq \mathbb R $$ One can see that this is a ring, and here 2 or 7 no longer are primes since $2=(\sqrt 2)^2$ and $7=(3+\sqrt2)(3-\sqrt2)$, but one can prove that 5 still is a prime.

Obviously you can factor a real number into any set of numbers whose product is that real number. As for PRIME factorizations I know of one generalization of that probably DOESN'T work well for real numbers:

If you consider prime factorizations as finite sequences of exponents of the first primes (up until the sequence ends), then positive integers represent the special case where these exponents are all non-negative integers:

$$ 2^3 3^2 5^0 7^1 = 504 $$

If you allow negative integers, then you get unique prime factorizations for all positive rational numbers:

$$ 2^3 3^{-2} 5^0 7^1 = \frac {56} {9}$$

If you allow any rational number exponents (positive, zero, or negative), then you get all rational POWERS of positive rational numbers:

$$ 2^{\frac {3} {2}} 3^{- \frac {2} {3}} 5^1 = \left(\frac {2560} {81}\right)^{\frac {1} {6}}$$

You can keep going by expanding the set of exponents; however, as soon as your set of exponents includes numbers of the type $ \log_p q $, where p and q are different prime numbers, then you lose uniqueness of prime factorizations:

$$ 2^{\log_2 {3}} = 2^0 3^1 = 3$$

so, if you just let any real number be an EXPONENT, you lose uniqueness. Technically, since each positive real number has at least one prime factorization, you can make a rule to choose which one is the primary one, and make a set of exponent sequences that only includes those.

In the context where I wanted these, is was natural to expect all these sets of numbers I wanted prime factorizations for to be closed under multiplication, which meant the sets of exponent sequences had to be closed under element-wise addition. You can trivially do this by choosing the primary prime factorization to be the one where only 2 has an exponent or similar, but those solutions are basically just logarithms, and, if the same rule were applied to integers, rational numbers, etc., they would no longer have the same prime factorizations as before, so this is not really an EXTENSION of prime factorizations. Quotient spaces provide a similar but slightly more satisfying solution.*

Just to state it out clearly, you can consider these sequences of exponent to infinite, but all zeros after a certain point, since there are infinite primes for them to be exponents of. You can also then view these as vectors in a countably infinite-dimensional space, if that makes the element-wise addition seem more natural, or if it helps you to see the conversion to the numbers they are prime factorizations of as using a dot product:

$$ r = b^{\vec M(r) \cdot [\log_b 2, \log_b 3, \log_b 5, ...]^T} $$

where $\vec M$ is the function that converts the number r to its "prime factorization", and b is any number that can be a base of a logarithm (I would be inclined to pick either b=2 or b=e as a default, or maybe $b=2^{1/12}$ for music theory or b=10 for some other kind of convenience, but it doesn't matter what you pick.) $\vec M$ is only a function if "prime factorizations" are unique of course.

However, if you allow infinite sequences of exponents that DON'T end in all zeros, you run into problems. Firstly, in the case where exponents have to be non-negative integers, any sequence with infinitely many non-zero exponents will lead to a product that diverges (so, in a sense, infinity clearly has no unique prime factorization). If you allow negative and/or rational exponents, then you can get products that actually converge, but you'll probably lose uniqueness, and also the set of possible products will be larger than with finite sequences of exponents. (I wouldn't be surprised if you can basic Analysis axioms and theorems and logarithms to prove that you can reach any positive real number this way with just negative exponents, or any real number greater than or equal to one with just positive rational exponents.)

If you want negative numbers, you can introduce -1 as a "prime number", but it behaves weirdly with loss of uniqueness and introduction of complex numbers if you allow it to take the same exponents as the actual primes. You can also use $i = \sqrt {-1} $ instead of -1 and get similar issues. I don't know what happens if you use Gaussian primes or "primes" of other rings, but it would probably work better than with using units like i and -1.

*I edited that paragraph after realizing you can technically preserve closure of the set of exponents under addition by making the set of allowable exponents a quotient space, but said quotient space is one-dimensional, and so just isomorphic to the real numbers, sort of in line with Qiaochu Yuan's answer:

Let $U = \{ (x_i)_{i=1}^{\infty} \in \mathbb{R}^{\mathbb{N}} | \sum_{i=1}^{\infty} x_i \log_b p_i = 0\}$. Then $\mathbb{R}^{\mathbb{N}}/U$ is a set of unique prime factorizaions for each number. For example, for the number 3,

$$\vec M(3) = [0, 1, 0, 0, 0, ...] + U$$

because (using $\vec L = [\log_b 2, \log_b 3, \log_b 5, ...]^T$)

$$b^{\vec L \cdot ([0, 1, 0, 0, 0, ...]^T + U)} = \{b^{x_1\log_b 2 + (1+x_2)\log_b 3 + x_3 \log_b 5 + ...} | (x_i)_{i=1}^{\infty} \in U\} = \{3\}$$

That is to say that the set of exponents sequences which map to $b^0 = 1$ (which is U) for a subspace of the vector space of infinite sequences of real numbers, so, for each positive real number r, the set of exponents sequences which map to r is a coset of U, a.k.a. an element of $\mathbb{R}^{\mathbb{N}}/U$, where $\mathbb{R}^{\mathbb{N}}$ is the set of all infinite sequences of real numbers (all maps from $\mathbb{N}$ to $\mathbb{R}$). Closure under addition uses the rule that:

$$\forall \vec a, \vec b \in \mathbb{R}^{\mathbb{N}}, (\vec a + U) + (\vec b + U) = (\vec a + \vec b) + U$$

Again, $\mathbb{R}^{\mathbb{N}}/U$ is one-dimensional and so isomoprphic to $\mathbb{R}$ in a way which makes finding the "prime factorization" coset of r not too different from taking a logarithm of r. Also, technically $\mathbb{R}^{\mathbb{N}}$ includes sequences that would map to infinity, and the set which does is not a coset of U. However, if you exclude sequences where $\sum_{i=0}^{\infty} \log_i p_i$ for primes $(p_i)_{i=1}^{\infty}$ is not absolutely convergent, that gives another vector space (a subspace of $\mathbb{R}^{\mathbb{N}}$), and even the set of only sequences that end in all zeros is a vector space, so replacing my "$\mathbb{R}^{\mathbb{N}}$" with either of those (starting with the definition of U) would make the above statements work.

Let's take a different perspective than the other answers have taken so far. You can think of prime factorizations as telling you exactly how the natural numbers $\mathbb{N}$ (here I am excluding zero) behave under multiplication; there are these primes $p$, and every natural number is uniquely a product of a bunch of primes. In abstract language, $\mathbb{N}$ under multiplication is a type of algebraic structure called a commutative monoid, and prime factorization tells us that $\mathbb{N}$ is the free commutative monoid on the primes.

So the primes tell us exactly how natural numbers multiply. One way to ask the question for real numbers is:

How do real numbers multiply?

That is, how can we understand the behavior of, say, the positive real numbers $\mathbb{R}_{+}$ as a commutative monoid?

The answer for the real numbers is quite different compared to the natural numbers: multiplication works the same way as addition! Formally, for any positive real $r > 1$ we can consider the exponential

$$\mathbb{R} \ni x \mapsto r^x \in \mathbb{R}_{+}$$

and this is an isomorphism of commutative monoids; what this means is that the familiar exponential law

$$r^{x + y} = r^x r^y,$$

together with the fact that the exponential has an inverse given by the logarithm, establishes that multiplication and addition of real numbers work exactly the same way; we can pass from addition to multiplication by taking the exponential, and we can pass from multiplication to exponentiation by taking the logarithm $\log_r$. This is a very fundamental fact, and before the advent of calculators you used to have to learn how to use slide rules to take advantage of this, and multiply complicated numbers by adding their logarithms.

So, we could say loosely that exponentials and logarithms establish that every positive real number $x$ has a "unique continuous factorization" into a "continuous product" of "$\log_r x$ copies" of any fixed real number $r > 1$. Taking $r$ to be $e$ would be a particularly nice choice but we don't have to.

Any rational $x>0$ can be uniquely written as $x=\prod{p_i}^{m_i}$ with $p_i$ the $i^\text{th}$ prime and $m_i\in\mathbb Z$ its associated multiplicity. This extends the notion of factorization to strictly positive rationals.

We can extend to non-zero rationals by writing $x=(-1)^{m_0}\prod{p_i}^{m_i}$ with $m_0\in\{0,1\}$. That's similar to how factorization is extended from $\mathbb N^*$ to $\mathbb Z^*$ in some contexts.

I don't see a meaningful extension to reals.