You raise a specific question the comments: Let $G$ be a finite group, let $k$ be an integer and let $\Gamma$ be the group on $k$ generators with relators given by the identities of $G$. Is $\Gamma$ finite for all $G$ and $k$?
I don't know the answer, but the question does at least reduce to the case when $G$ is a finite simple group. To keep things short I will use 'identity of $G$' to mean a word whose only value on $G$ is the identity element. Suppose that $G$ has minimal order among the finite groups for which $\Gamma$ is infinite (for $k$ large), and that $G$ is not simple. Let $K$ be a normal subgroup of $G$ such that $G/K$ is simple, and let $\Gamma_1$ be the subgroup of $\Gamma$ generated by all identities of $G/K$. Then $\Gamma/\Gamma_1$ is finite by the minimality of $G$, so $\Gamma_1$ is finitely generated; let $X$ be some finite generating set for $\Gamma_1$. Then by the minimality of $G$, $\Gamma_1/\Gamma_2$ is finite, where $\Gamma_2$ is generated by all the identities of $K$, written in the alphabet $X$. I claim that $\Gamma_2$ is actually trivial: if we write an identity of $K$ using letters that are themselves identities of $G/K$, we get an identity of $G$. Thus $\Gamma$ is finite, contradiction.
The answer is clearly 'yes' for the cyclic group of order $p$ (and hence for all finite soluble groups): in this case $\Gamma$ is elementary abelian of order $p^k$. For the non-abelian finite simple groups, I can't see a clever way of doing it - it would be nice to see a proof that doesn't involve a CFSG trawl, certainly.