Perhaps someone can explain this category theoretic fact. I must have read somewhere that it follows from the fact that there's a left adjoint for this functor. That would be the group ring construction.
So I guess my question boils down to: why does the existence of a left adjoint for a functor guarantee that it respects (preserves) products?
Here's a low-tech answer before answering about adjoints (I'm not sure whether that was part of your question, or whether your question was purely about the category-theoretic proof, but let me do it anyway for potential passer-by's):
a unit in $X\times Y$ is a couple $(x,y)$ with an inverse, that is, a couple $(x',y')$ with $(x,y)(x',y') = (1,1)$. Since multiplication is defined pointwise, this is equivalent to $xx' = 1, yy' = 1$ (and symmetrically for $x'x, y'y$) so that if $x,y$ are invertible, so is $(x,y)$.
Now here's the category theoretic version (this is a summary from the comments essentially): $R\mapsto R^\times$ is right adjoint to $G\mapsto \mathbb Z[G]$ as is easily checked, so that it must commute with all limits : this is a general fact, right adjoints always preserve limits.
Moreover, products are limits: $A\times B$ is the limit of the functor from the category $\bullet \space\bullet$ with values $A$ and $B$
The proof in that particular case could go as follows: $\hom(G,(X\times Y)^\times) \cong\hom(\mathbb Z[G],X\times Y)\cong \hom(\mathbb Z[G],X)\times \hom(\mathbb Z[G],Y)\cong \hom(G,X^\times)\times \hom(G,Y^\times) \cong \hom(G,X^\times\times Y^\times)$ and so $(X\times Y)^\times \cong X^\times \times Y^\times$ by the Yoneda lemma.
