In the presence of the axioms for a commutative monoid, idempotency is equivalent to self-distributivity.
Proof. Suppose a commutative monoid is idempotent. Then: $$x(yz) = xxyz = (xy)(xz)$$
On the other hand, suppose a monoid is self-distributive on the left. Then
$$x = x(11) = (x1)(x1) = xx$$
Question. What are some examples of non-commutative monoids that are both idempotent and self-distributive (on both sides)?
(We seem to agree that a monoid is associative and has a unit element.) Left self-distributivity implies that for all $x$ and $y$ we have $xy = (xy)1 = x(y1) = (xy)(x1) = (xy)x = xyx$. But right self-distributivity symmetrically implies $yx = xyx$. So $xy = yx$ for any $x$ and $y$, and the monoid is commutative.
I hope I'm not stating the obvious, but in general, to think about such questions, the best method seems to be to start with the free monoid on some alphabet and impose the desired properties by quotienting out by the appropriate word relations: for example, idempotency means we identify $uxv$ with $uxxv$ for any words $u,x,v$, left-distributivity means we identify $uxyzv$ with $uxyxzv$ for any $u,x,y,z,v$, and so on. The question is then to find words which cannot be related (here $xy$ and $yx$) by these rewriting rules. Of course, one must remember the empty word, but thinking in terms of words makes the problem typically easier.