Here is how I think I proved that any group of cardinality $4$ is abelian:
Claim: a group in which any element $a\ne e$ has order $2$ is abelian
Proof: Let $G$ be such a group and let $a,b\ne e$ be in $G$. Since $G$ is a group $ab\in G$ and $(ab)\cdot(ab)=e \text{ because }g^2=e\iff g^{-1}=g$
$\implies aba^{-1}b^{-1}=e$
Multiplying by $b$ on the right $aba^{-1}=b$ and since $a,b$ were arbitrary we get the desired result by multiplying by $a$ on the right: $ab=ba$
Proof that $\#G=4\implies G$ is abelian:
Let $a\in G\backslash\{e\}$ be an arbitrary element. By Lagrange the order($a$) divides 4. If order$(a)=2$ let's choose $b\ne a$ and $b\ne e$. By the same reasoning either order$(b)=2$ or $4$. If it's $4$ we're done, if it's $2$ let's choose $c\ne e,a,b$. Now if the order of $c$ is $2$ we can use the claim to finish the proof.
If order$(a)=4$ then $G=\langle a\rangle$ is cyclic and so isomorphic to $\Bbb Z/4\Bbb Z$ so abelian.
Without any appeal to orders of elements, Cauchy's Theorem or Lagrange's or type of group: assume that $G$ has $4$ elements and is not abelian. Then we can find two non-identity elements $a,b$ that do not commute, so $ab \neq ba$. Note that this implies $ab \notin \{e,a,b\}$. Because $G$ is closed under the group operation we must have $G=\{e,a,b,ab\}$. Now $ba$ belongs to this set. Since $a$ and $b$ do not commute $ba \notin \{e,a,b\}$. Hence we must have $ab=ba$, a contradiction. So $G$ is abelian after all.