Is there an explicit example of a non-commutative monoid $M$ such that for all elements $m,n \in M$ and $p \in \mathbb{N}$ we have $(m \cdot n)^p=m^p \cdot n^p$?
It suffices to construct a semigroup $H$ with an absorbing element $0$ such that $a^2=0$ for all $a$, because then $M := H \cup \{e\}$ will do the job. This sounds easy at first glance, but I cannot find an example.