I'm trying to a build a better understanding of presentations. I get that a a group has a presentation $\langle S \mid R \rangle$ if it is the "freest" group subject to the relations $R$.
But, for instance, how is it immediately clear that $\langle x, y \mid x^2 = y^2 \rangle$ is the same group as $\langle x, y | xyx^{-1}y \rangle$? Also, more generally, what are some ways to determine if two different presentations yield the same group?
On
Sometimes the laws of the presentation imply eachother. But in general, you can think of them as quotient groups of the free groups, and the relations as congruence relations, or with manipulation, generators of the kernel (which is why the default relation is to the identity). And then it isn't usually immediately apparent what things are and aren't isomorphic, and the presentation may not all-knowing. So you prove the isomorphisms theorems.
On
It is a little difficult to try to explain this without drawing stuff, but if you draw a rectangle with one pair I of opposite sides identified and the other pair II identified in contrary directions, you can see you have two Moebius bands: one as a strip "in the middle" of the rectangle with the sides II identified, and the other one taking the strips above and before this first Möbius band (I'm sure there must be some site where this thing's done!).
Now you can apply the Seifert- van Kampen Theorem by joining the two bands on the middle of the rectangle, one above I and the other one below it, and this renders you the group $\,\Bbb Z *_{\Bbb Z}\Bbb Z\,$ , when the amalgation gives us just the relation $\,x^2=y^2\,$.
If you succeed in grabbing the geometric picture above you have your isomorphism at once, otherwise it is going to take way more time and sweat. If I can come up with something later I'll add it here.
Added: Massey's IV.5 seems to work on this but it is too complicated for me to understand it by merely reading it quickly. If you have some time try it.
Further added: We have $\,x^2=y^2\Longrightarrow x^2y^{-2}=1\,$ , and thus we can put
$$r:=xyx^{-1}\;,\;\;s:=xy^{-1}\Longrightarrow$$
$$ rsr^{-1}s=xyx^{-1}xy^{-1}xy^{-1}x^{-1}xy^{-1}=x^2y^{-2}=1$$
and this way we can pass from your first presentation to the second one...
As user58512 points out in the comments, it is not always possible to determine whether two presentations define isomorphic groups. It has been shown that the general problem of determining whether a given presentation defines the trivial group is undecidable.
However, in many cases we can show (or even easily see) that two presentations represent the same group. One way to do this formally is using Tietze transformations. I suggest you try transforming your first presentation into the second using Tietze tranformations. As a hint, try starting by adding $xy$ as a new generator in the second presentation.