Order of nontrivial elements is 2 implies Abelian group

Taking inverses reverses the order of multiplication, so if every element is its own inverse multiplication must be commutative.

As every non-identity element has order two, $a^{-1} = a$ for any element of the group. Therefore $$[a, b] = aba^{-1}b^{-1} = abab = (ab)^2 = e.$$ Hence the group is abelian. Is this too calculationy?

$[a,b]=1$ for all $a,b\in G$ if and only if $G$ is abelian. You proved that $[a,b]=a^{-1}b^{-1}ab=1$ above - this is the connection to commutators.

I don't know of any strong motivation behind this fact aside from, I guess, knowing that any nonabelian group must have an element of order $>2$. I think that it is just a standard exercise.

It may interest you motivationally to prove that $G/H$ is abelian if and only if $G'\leqslant H$ (if $H \unlhd G$).