If every nonidentity element in a group is of order $2$, the group is abelian

Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^{-1}=ab$, but $(ab)^{-1}=b^{-1}a^{-1}=ba$ by using twice again the remark.

Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).