If $G$ is abelian, then the set of all $g \in G$ such that $g = g^{-1}$ is a subgroup of $G$

A different way to phrase the same argument everyone gave:

The map $a\in G\mapsto a^2\in G$ is a group homomorphism and your subset $H$ is its kernel: it is therefore a subgroup of $G$.

$\newcommand{\N}{\Bbb N}$ Let $G$ an abelian group, let $e$ denote its identity element. For each $m\in\N$ define $$G(m):=\{g\in G: g^m=e\}.$$ $G(m)$ is a subgroup of $G$. Indeed, you can see that $e\in G(m)$. If $g,h\in G$, since $G$ is abelian we have $$(gh^{-1})^m=g^m(h^{-1})^m=e(h^m)^{-1}=e^{-1}=e.$$ Therefore $G(m)\leq G$ as claimed.

You'll need to show only closure under multiplication (that is, that $ab\in H$ for all $a,b\in H$), since the identity is trivially its own inverse, so is in $H$, and since every element of $H$ is its own inverse, you don't need to check inverses, either. The fact that $G$ (so also $H$) is abelian makes checking closure fairly trivial.

Generally the one-step subgroup test is faster but in this case you can just check the group axioms: the only non-trivial one is closure. If $a^2=b^2=e$, can you see that $ab$ is its own inverse, given the group is Abelian?