Ultracoproducts and Cartesian products

Aaron, the answer is big no if $X$ and $Y$ are infinite (no metrasibility assumptions required to show that the answer is always negative).

Indeed, $C((X\times Y)^{\mathcal{U}})$ is a Grothendieck space as an ultraproduct of $\mathscr{L}_\infty$-spaces over a countably incomplete ultrafilter. (See Propositon 3.2 here.) On the other hand $$C(X^{\mathcal{U}}\times Y^{\mathcal{U}})\cong C(X^{\mathcal{U}})\otimes C(Y^{\mathcal{U}})\cong C(X^{\mathcal{U}}, C(Y^{\mathcal{U}})) $$ contains a complemented (Banach-space) copy of $c_0$ by the main result of

P. Cembranos, $C(K, E)$ contains a complemented copy of $c_0$, Proc. Amer. Math. Soc. 91 (1984), 556-558

so it is manifestly not Grothendieck. The conclusion is that $C((X\times Y)^{\mathcal{U}})$ and $C(X^{\mathcal{U}}\times X^{\mathcal{U}})$ are not even Banach-space isomorphic.

Remark. One can tweak Cembranos' proof to obtain the following result:

If $A$ is a C*-algebra, $E$ is a Banach space (both $A$ and $E$ are infinite dimensional) and $\gamma$ is any Banach-space tensor cross-norm on $A\odot E$, then $A\otimes_\gamma E$ is not a Grothendieck space.

As a corollary to this one obtains a negative answer to Simon Wasserman's quesiton whether the Calkin algebra can be written as a non-trivial tensor product of two C*-algebras. I might write this down at some point.