As an alternative approach with the same solution:

Given $g\in G$, I will show that $g$ is of order $2$ (or $1$). From here on, a subset of $G$ is "large" if its complement is of lower cardinality than that of $G$.

Let $G^{(2)}$ be the set of elements of order $2$. Then $G^{(2)}z$ is large, so $G^{(2)} \cap G^{(2)}z$ is large. We can therefore choose $x$ such that $x, xg$ are both of order $2$.

Similarly, we can find $y$ such that $y, z:= yxg, xy, xz = xyxg$ are all of order $2$. Then by the usual proof, $x, y, z$ all pairwise commute and are of order $2$, so $xyz = xyyxg = xxg = g$ is of order $2$ (or $1$).

Interestingly, this approach should be easy to adapt to measure theory, leading to a conclusion of the form: Let $G$ be a group with a finite biinvariant measure (and assume the measure is normalized to $1$). Then if the set of elements of order $2$ is measurable with measure larger than $c$, the group is commutative and every element has order $2$. I think this proof demonstrates the $c=\frac{5}{6}$ statement, and that $c=\frac{1}{2}$ is likely.

As an alternative approach with the same solution:

Given $g\in G$, I will show that $g$ is of order $2$ (or $1$). From here on, a subset of $G$ is "large" if its complement is of lower cardinality than that of $G$.

Let $G^{(2)}$ be the set of elements of order $2$. Then $G^{(2)}z$ is large, so $G^{(2)} \cap G^{(2)}z$ is large. We can therefore choose $x$ such that $x, xg$ are both of order $2$.

Similarly, we can find $y$ such that $y, z:= yxg, xy, xz = xyxg$ are all of order $2$. Then by the usual proof, $x, y, z$ all pairwise commute and are of order $2$, so $xyz = xyyxg = xxg = g$ is of order $2$ (or $1$).

Interestingly, this approach should be easy to adapt to measure theory, leading to a conclusion of the form: Let $G$ be a group with a finite one-sided invariant finitely-additive measure (and assume the measure is normalized to $1$). Then if the set of elements of order $2$ is measurable with measure larger than $c$, the group is commutative and every element has order $2$. The limiting step in the above proof is the choice of $y$, and there are $4$ conditions on $y$, so $c = \frac{3}{4}$ should work. Considering the dihedral group of the square, in which $6$ out of $8$ elements are of order $2$ or less, $\frac{3}{4}$ should be optimal.

As an alternative approach with the same solution:

Given $g\in G$, I will show that $g$ is of order $2$ (or $1$). From here on, a subset of $G$ is "large" if its complement is of lower cardinality than that of $G$.

Let $G^{(2)}$ be the set of elements of order $2$. Then $G^{(2)}z$ is large, so $G^{(2)} \cap G^{(2)}z$ is large. We can therefore choose $x$ such that $x, xg$ are both of order $2$.

Similarly, we can find $y$ such that $y, z:= yxg, xy, xz = xyxg$ are all of order $2$. Then by the usual proof, $x, y, z$ all pairwise commute and are of order $2$, so $xyz = xyyxg = xxg = g$ is of order $2$ (or $1$).

As an alternative approach with the same solution:

Given $g\in G$, I will show that $g$ is of order $2$ (or $1$). From here on, a subset of $G$ is "large" if its complement is of lower cardinality than that of $G$.

Let $G^{(2)}$ be the set of elements of order $2$. Then $G^{(2)}z$ is large, so $G^{(2)} \cap G^{(2)}z$ is large. We can therefore choose $x$ such that $x, xg$ are both of order $2$.

Similarly, we can find $y$ such that $y, z:= yxg, xy, xz = xyxg$ are all of order $2$. Then by the usual proof, $x, y, z$ all pairwise commute and are of order $2$, so $xyz = xyyxg = xxg = g$ is of order $2$ (or $1$).

Interestingly, this approach should be easy to adapt to measure theory, leading to a conclusion of the form: Let $G$ be a group with a finite biinvariant measure (and assume the measure is normalized to $1$). Then if the set of elements of order $2$ is measurable with measure larger than $c$, the group is commutative and every element has order $2$. I think this proof demonstrates the $c=\frac{5}{6}$ statement, and that $c=\frac{1}{2}$ is likely.