This answer is somewhat tangential to the original question, and follows the direction suggested by Sean Eberhard's comment.

For a finite group $G$, it is the case that if more than $\sqrt{\frac{5}{8}} |G|$ elements $x \in G$ have $x^{2} = e$, then $G$ is Abelian. The dihedral group of order $8$ ( I mean the one with $8$ elements) - and direct products of it with elementary Abelian $2$-groups as large as you like-show that this can't be improved much as a general bound, since a dihedral group $D$ of order $8$ contains $6$ elements which square to the identity and $ 6 < \sqrt{\frac{5}{8}} |D| <7$ in that case.

This is because (as noted in the paper linked to in Sean Eberhard's comment, and also previously noted by Brauer and Fowler), the count of solutions to $x^{2} = e$ given using the Frobenius-Schur indicator leads easily to $\sqrt{\frac{5}{8}} |G| < \sqrt{k(G)}\sqrt{|G|}$ in the case under consideration, where $k(G)$ is the number of conjugacy classes of $G$. Hence $\frac{k(G)}{|G|} > \frac{5}{8}$, so the probability that two elements of $G$ commute is greater than $\frac{5}{8}$, in which case $G$ is Abelian by a Theorem of W.Gustafson. If preferred, this can be seen directly using character theory- in general, if $G$ has $k$ conjugacy classes and $Z = Z(G)$, we have $|Z| + 4(k - |Z|) \leq |G|$ by the orthogonality relations. Hence $\frac{k}{|G|} \leq \frac{1}{4} + \frac{3|Z|}{4|G|} \leq \frac{5}{8}$ if $Z < G$.