What is the Cayley graph of $(\Bbb Z/2\Bbb Z)\times(\Bbb Z/2\Bbb Z)$? I get that the presentation of the new group, with respect to two generators, would be $(x,y \;|\; x^2= y^2=1)$ but I'm not sure how the actual graph would look. Would it consist of an infinite sequence of loops, where each loop has one edge for the $x$ generator and one for the $y$? But then how would the $x^2$ relations come into play? Any help in my understanding would be greatly appreciated. Thanks!

No it's not infinite. You have the four points $1, x, y, xy$ and with $S=\{x,y\}$ as your generating set, you get edges $\{1,x\}, \{1,y\}, \{x,xy\}, \{y, xy\}$.
Also your presentation is incorrect, what you have written is $\Bbb Z/2\Bbb Z * \Bbb Z/2\Bbb Z$, you need to add the relation $xyx^{-1}y^{-1}=1$, otherwise there are infinitely many nodes.
Note each node has degree $2$, which is exactly right since $|S|=2$. So the graph looks like a square with $1$ in the upper left, $x,y$ adjacent to it, and $xy$ in the lower right.
Remember you only get one edge per generator. You do not get edges for anything you can get to eventually, otherwise your graph would always be a complete graph by the definition of a generating set.
