If the order of all nontrivial elements in a group is 2, then the group is Abelian. I know of a proof that is just from calculations (see below). I'm wondering if there is any theory or motivation behind this fact. Perhaps to do with commutators?
Proof: $a \cdot b = (b \cdot b) \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.
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 $H \unlhd G$).
The idea of this approach is to work with a class of very small, finite, subgroups $H$ of $G$ in which we can prove commutativity. The reason for this is to be able to use the results like Cauchy's theorem and Lagrange's theorem.
Consider the subgroup $H$ generated by two distinct, nonidentity elements $a,b$ in the given group. The group $H$ consists of strings of instances of $a$ and $b$. By induction on the length of a string, one can show that any string of length 4 or longer is equal to a string of length 3 or shorter.
Using this fact we can list the seven possible elements of $H$: $$1,a,b,ab,ba,aba,bab.$$ By (the contrapositive of) Cauchy's Theorem, the only prime divisor of $|H|$ is 2. This implies the order of $H$ is either $1$, $2$, or $4$.
If $|H|=1$ or $2$, then either $a$ or $b$ is the identity, a contradiction.
Hence $|H|$ has four elements. The subgroup generated by $a$ has order 2; its index in $H$ is 2, so it is a normal subgroup. Thus, the left coset $\{b,ba\}$ is the same as the right coset$\{b,ab\}$, and as a result $ab=ba$.
Lemma: Let $G$ be a group. Let $g\in G\setminus \{e\}$. Then $|g|=2$ if and only if $g=g^{-1}.$
Proof: Suppose $|g|=2$. Then $g^2=e$, so
$$\begin{align} g&=ge\\ &=g(gg^{-1})\\ &=g^2g^{-1}\\ &=eg^{-1}\\ &=g^{-1}. \end{align}$$
Now suppose $g=g^{-1}$. Then $g^2=gg=gg^{-1}=e$. But $g\neq e$. Hence $|g|=2.$ $\square$
Lemma: Let $g,h$ be in a group $G$. Then $$(gh)^{-1}=h^{-1}g^{-1}.$$
Proof: We have that inverses are unique and $$(gh)(h^{-1}g^{-1})=geg^{-1}=gg^{-1}=e.\, \square $$
Theorem: Suppose $a^2=e$ for all $a\in G$, where $G$ is a group. Then $G$ is abelian.
Proof: Let $g,h\in G$. Then
$$\begin{align} gh&=(gh)^{-1}\\ &=h^{-1}g^{-1}\\ &=hg. \end{align}$$
But $g,h\in G$ were arbitrary. Hence $G$ is abelian. $\square$
