Let $X$ be a metrizable compact topological space, let $\mathcal U$ be an ultrafilter, and denote by $X^{\mathcal U}$ the ultracopower of $X$ with respect to $\mathcal U$. As a C$^*$-algebraist, I prefer to define $X^{\mathcal U}$ as the compact Hausdorff space satisfying \begin{align*} C(X^{\mathcal U},\mathbb C) &\cong \prod_{\mathcal U} C(X,\mathbb C) \\ &:= \ell^\infty(\mathbb N, C(X, \mathbb C))/\{(f_n)_n \mid \lim_{n\to\mathcal U} \|f_n\|_{\infty} = 0\}. \end{align*} It can alternatively be described, directly, as ultrafilters in $\mathbb N \times X$ which respect $\mathcal U$ (see Bankston's "Reduced coproducts of compact Hausdorff spaces" for a more precise version of this).

My question is: if $Y$ is another metrizable compact topological space, is it the case that $(X \times Y)^{\mathcal U} \cong X^{\mathcal U} \times Y^{\mathcal U}$? Equivalently, in terms of C$^*$-algebras, is it the case that $$ \prod_{\mathcal U} C(X \times Y, \mathbb C) \cong \left(\prod_{\mathcal U}C(X, \mathbb C)\right) \otimes \left(\prod_{\mathcal U} C(Y, \mathbb C)\right)? $$

My intuition, looking at the C$^*$-algebra version, is that this isn't true, although I've found it difficult to prove this.

(Note: we could do an ultracoproduct with respect to a sequence of spaces instead, but this is notationally messier and, well, let's first see the answer to the question I stated before going there.)