Examples of rings whose units $\cup\, \{0\}$ comprise a field

We have that $R$ satisfies your property if and only if $k = R^\times\cup \{0\}$ is a field. So we can state this problem inversely: if $k$ is a field, for which ring extensions $k\subset R$ is it the case that $R^\times = k^\times$?

A first observation: if $\operatorname{char} k\neq 2$, then every element of $R$ must be transcendental over $k$. This is because, if $x\in R$ is algebraic, then $k[x]$ is a product of field extensions of $k$. Each field extension must be trivial, to avoid increasing the unit group, and there can only be one field in the product, for the same reason (here we use $\operatorname{char} k\neq 2$). So $x\in k$.

Now, if we have $R^\times = k^\times$, then $S^\times = k^\times$ for any $k\subset S\subset R$. So we should begin our search by considering the case where $R$ is finitely generated over $k$. Said differently: we know that $R$ is a quotient of $k[X_\alpha]$ for some indexing set, but we are interested in what relations can hold between the variables.

This problem now has a nice statement in algebraic geometry: for what affine varieties $X$ over $k$ do we have $k[X]^\times = k^\times$? Equivalently: when are all non-vanishing regular functions on $X$ constant?

A simple example is zcn's $k[x,y]/(xy)$, the coordinate ring of two intersecting lines, which is a coolring for any $k$. guy-in-seoul just put this in his answer, so see his post for details. In short, joining two coolvarieties at a point will yield another coolvariety, so two lines is the natural non-polynomial-ring example.

As another example, if $X$ is a non-singular cubic plane curve given by $y^2 = x^3 + ax+b$ (or any affine curve with only one point at infinity), then all non-vanishing regular functions on $X$ are constant. For if $X\to\mathbb{A}^1_k \subset \mathbb{P}^1_k$ is a morphism, then it can be extended to $\tilde{X} = X\cup\{\infty\}\to \mathbb{P}^1_k$, which must be surjective (impossible) or constant.

Any ring with trivial unit group is a coolring: if $1$ is the only unit, then the ring has characteristic $2$ (since $-1 \in R^\times \implies 1 = -1$), so the only sum of two units is $1 + 1 = 0$. The semilocal rings with trivial unit group are precisely finite products of $\mathbb{Z}/2\mathbb{Z}$, and if the product has more than one factor, these are not of the form $R[t]$ for any ring $R$ (since every nonunit is a zerodivisor).

For an infinite counterexample, take $S := (\mathbb{Z}/2\mathbb{Z})[x,y]/(xy)$, which has trivial unit group. If it were of the form $R[t]$ for some ring $R$, then $R$ would be Artinian with trivial unit group, hence $R = \prod_{i=1}^n \mathbb{Z}/2\mathbb{Z}$ is a finite product of $\mathbb{Z}/2\mathbb{Z}$'s. Then $n > 1$ since $S$ is not a domain, so $R[t]$ has idempotents, but $S$ does not (for proofs of some of these statements, see this answer).

Update: As other answers have given some geometric examples, let me mention a truly abundant source: any homogeneous coordinate ring of a projective variety is a coolring (over any field $k$). Indeed, such a ring is of the form $R = k[x_0, \ldots, x_n]/P$ for some homogeneous prime $P$, so the projection $k[x_0, \ldots, x_n] \twoheadrightarrow R$ induces a surjection $k^\times = k[x_0, \ldots, x_n]^\times \twoheadrightarrow R^\times$ (since $R$ is a postively graded domain, so $R^\times = (R_0)^\times$). Generalizing, a ring of the form $k[x_0, \ldots, x_n]/J$, where $J$ is a homogeneous radical ideal (and thus a finite intersection of homogeneous primes), is a coolring.