# Is an associative binary operation with trivial squares necessarily commutative?

Take a set $$S$$ and an associative binary operation $$*:S \times S \rightarrow S$$ such that there exists an element $$e$$ such that $$x * x = e$$ for any $$x \in S$$. Can we conclude that the operation is commutative?

This question was inspired while trying to prove easy algebra statements with SPASS. The original problem supposed that $$S$$ was a group and $$e$$ the identity, but I tried to remove hypotheses and found out that the statement remains true if we simply suppose that $$e$$ is a left/right identity for the binary operation (still assuming that this operation is associative). Trying to remove this hypothesis made the saturation process diverge. I suppose that a counterexample exists and probably SPASS is not running with optimal settings, but I don't know how to find one.

$$\begin{array} {r|r r r r r} * & a & b & c & d \\ \hline a & a & a & a & a \\ b & a & a & a & a \\ c & a & d & a & a \\ d & a & a & a & a \\ \end{array}$$
The hardest part is checking associativity, but a little thought reveals that $$(x * y) * z = x * (y * z)$$ is always $$a$$ (think about what $$x * y$$ can be and what that tells you about $$(x * y) * z$$).