Does there exist a non-commutative, non-pure, non-group monoid such that its set of invertible elements commutes with every element?

Question

Does there exist a non-commutative monoid $(M,*)$, which is not a group, and which is also not pure, meaning it has at least one non-identity invertible element, such that the set of invertible elements $U$ of $M$ commutes under multiplication with every element $a$ of $M$, in symbols: $a*U = U*a$? Of course, if $M$ is commutative, or a group, or pure, then certainly $U$ commutes under multiplication with every element $a$ of $M$. Hence, I am looking for a non-trivial example of that phenomenon. This question was inspired by a text I read online.

Answer 1

Take the direct product $A \times N$ of a nontrivial abelian group $A$ and a noncommutative pure monoid $N$, e.g. $\mathbb{Z}$ times the free monoid on $2$ generators. By construction the group of units is $A$ and by construction it commutes (even pointwise, which is stronger than what you asked for) with everything.