Existence of monoidable submagma of monoid which is neither a submonoid nor idempotent?

Question

Does there exist a monoid $(M;*,1)$ and a submagma $S$ of $M$, such that $S$ is not a submonoid of $M$, but $S$ is "monoidable", meaning it contains an identity element, and also, $S$ is not idempotent either? The reason I am asking this question is because I was considering the monoid $(N;*,1)$ of nonnegative integers under multiplication, and its monoidable submagma yet non-submonoid $\{0\}$. That example does not quite satisfy my condition, because it is idempotent. I am now wondering whether there are any non-idempotent examples of what I am looking for.

Answer 1

Let $M_1$ and $M_2$ be two monoids such that $M_1$ has a non-identity idempotent element $e$, and $M_2$ is not idempotent. Consider $M_1 \times M_2$. Then $\{e\} \times M_2$ is a subsemigroup but not a submonoid. It's isomorphic to $M_2$ as a semigroup though, so it's monoidable, if I've understood correctly (but not idempotent).

This means there is an example where $M$ is of order $4$, for example, as well as many more familiar examples like $\Bbb Z \times \Bbb Z$ under multiplication. An instance of this "in nature" is the monoid of $2 \times 2$ matrices over your favourite ring (say $\Bbb Z$ or $\Bbb R$) under multiplication, and the subsemigroup of matrices whose only nonzero entry is the top left one.

Answer 2

A 3-element monoid suffices. Take the monoid $M$ presented by $\langle a \mid a^3=a\rangle$. Then $M=\{1,a,a^2\}$ and $S = \{a,a^2\}$ is a non-idempotent subsemigroup of $M$. But $S$ is also a monoid, in fact a cyclic group of order 2, with identity $a^2$, since $a^2a^2=a^3a =aa$.