Is this modest change enough to make an idempotent element an identity element?

Question

In The Number System by Thurston the author introduces an algebraic structure he calls a hemigroup. The laws of a hemigroup are:

• (i) $\left(x\odot y\right)\odot z=x\odot \left(y\odot z\right),$
• (ii) $\left(x\odot y\right)=\left(y\odot x\right),$
• (iii) $\left(x\odot y\right)=\left(x\odot z\right)\implies{y=z},$
• (iv) $\exists_e e\odot e=e.$

I have been told (see comments) that $e$ of (iv) is not necessarily an identity element, but merely an idempotent element. I'm trying to understand what this actually means. I assume the distinction would be that $x\odot e=e\odot x=e\odot x\odot e,$ etc., but we have no rule $x\odot e=x.$ Please correct me if this is incorrect.

My main question is this: if $\mathcal{H}$ is the set of the hemigroup and we define our set of interest to be $\mathcal{S}\equiv \{\xi\backepsilon \xi=x\odot e\land e,x\in \mathcal{H}\},$ preserving the meaning of $\odot$ in $\mathcal{S},$ does $e$ now constitute an identity element of $\mathcal{S}?$ I am accepting the identity $e=e\odot e\in\mathcal{S}$.

Answer 1

Actually, $e$ is an identity element of $\mathcal{H}$ itself, because of axiom (iii). For any $x\in\mathcal{H}$, $$e\odot x=(e\odot e)\odot x=e\odot(e\odot x)$$ which by axiom (iii) implies $x=e\odot x$.

Answer 2

Yes, $e$ ($=e\odot e$) is a right identity in $\mathcal S$ because $$(x\odot e) \odot e = x\odot (e\odot e)= x\odot e$$ (and is a two sided identity because $\mathcal H$ was abelian to begin with).