No, I’ve never found a reference. I had even considered submitting a short note somewhere (perhaps to the AMM?), but then I got sidetracked by other things and forgot about it.

Here is the link to the question alluded to in my previous comment: mathoverflow.net/q/480608/16537

What about a reference for the existence of two infinite non-isomorphic semigroups whose finitary power semigroups are isomorphic? I was not aware of this result.

In fact, all what you need in this proof is seemingly that $\Phi$ is a surjective, non-injective endomorphism of a group $G$ and $N$ is the semidirect product of $G$ by the additive monoid $\mathbb N$ of non-negative integers wrt the left unitary action of $\mathbb N$ on $G$ given by $(m,a) \mapsto \Phi^m(a)$. AFAICS, it's not even necessary to assume that $G$ is abelian.

[...] Lastly, we look at the semidirect product $N := A \rtimes M$ of $A$ by $M$ relative to the given action, which is the Cartesian product $A \times M$ endowed with the binary multiplication defined by $$(x, m) (y, n) := (x + \Phi^m(y), m+n), \qquad\text{ for all }x, y \in A\text{ and }m, n \in \mathbb N.$$ The bottom line is that $N$ is a right cancellative, duo monoid that is not left cancellative.

(A note for my future self.) In this answer, $A$ is, up to iso, the direct sum of countably infinite copies of the additive grp of ints, and $M$ is the additive monoid of non-negative ints. Next, we consider the (left) (unitary) action of $M$ on $A$ that maps a pair $(n,x)\in M\times A$ to $\Phi^n(x)$, where $\Phi^n$ is the $n$-th iterate of the surjective endo of $A$ sending an element $x=(a_0,a_1,\ldots) \in A$ (uniquely represented by the integer coordinates $a_i$ relative to the canonical basis of $A$ as a free left $\mathbb Z$-module) to the element $(a_1,a_2,\ldots)\in A$. [...]

Again in the proof of Claim 1: shouldn't we also guarantee that multiplication by $s'$ Is injective when restricted to $S \setminus \{e,s,s^2\}$? And regarding the statement: don't we know for sure that $s \ne s^2$ given that $H_s$ is a non-trivial group?

It seems to me that, in the proof of Claim 1, you're implicitly assuming that $e \ne s$: I guess you want $H_s$ to be a non-trivial group in the statement, don't you? (Btw, why not writing $ez = s(s'z)$ rather than $ez = s'(sz)$? Both are of course correct, but the former emphasizes that, in the notation of the OP, you're taking $a = e$ and $b = s$. At first, I was confused by the notation and thought you were considering $a = e$ and $b = s'$.)