If we think of a U (set union) as a function what would be its domain and codomain?

As you have said, we want $\cup$ to be a function that takes as input two sets and returns another set. First off though, arguing with the universal set is tricky. What I would rather do is restrict yourself to a smaller set of possible inputs. For example, consider only finite sets, or only countable sets, ... In general, choose a particular set $X$ and then one can talk about such function, which takes subsets $A, B \subseteq X$ and returns another subset $C \subseteq X$.

Now the domain of such a function should be the set containing all pairs of subsets of $X$ i.e. the cartesian product of $P(X)$ with itself. Since the function returns a subset of $X$ the co domain should be $P(X)$. Hence $\cup: P(X) \times P(X) \rightarrow P(X)$.

Since $\emptyset \subseteq A$ we have $\emptyset \in P(X)$. By the definition of the cartesian product of sets $(\emptyset, A)$ and $(A, \emptyset)$ are two distinct elements in $P(X) \times P(X)$. However, since $A \cup \emptyset = A = \emptyset \cup A$ for any $A \in P(X)$ the function is not injective i.e. not 1-1.

Finally, for any set $A \in P(X)$ we have $A = \emptyset \cup A$. Hence $A$ is the image of $(\emptyset, A)$ under the function. This makes the function by definition surjective i.e. onto.

Apart from $A\cup B=\{\,x\mid x\in A\lor x\in B\,\}$ we can also consider the more general union $\bigcup A=\{\,x\mid \exists y\colon x\in y\in A\,\}$ (so if $A$ is a finite set of sets, $A=\{A_1,A_2,\ldots, A_n\}$, we have $\bigcup A=A_1\cup A_2\cup \ldots \cup A_n$).

Then $\bigcup$ can take as input any set ad produces an arbitrary set as output. Since the the class of all sets is not a set, speaking of domain set and range set is a bit problematic. The way in which $\bigcup$ is function-esque is that it is a class function. This just means that we have a predicate $ \Phi(x,y)$ with the properties $\forall x \exists y\colon \Phi(x,y)$ and $\forall x \forall y \forall z\colon \Phi(x,y)\land \Phi(x,z)\to y=z$ (in other words, for every set $x$ there exists one and only one $y$ such that $\Phi(x,y)$; we interpret $\Phi(x,y)$ as "$x$ is mapped to $y$"). The $\Phi$ for $\bigcup$ is readily written down: $$\Phi(x,y)\equiv \forall z\colon ( z\in y\leftrightarrow \exists w\colon z\in w\land w\in x).$$