Does a statement equivalent to Theorem 3 below appear in the literature? If it does, what is the earliest published reference?

Theorem 1. Let $W$ be a non-trivial cancellative invertible-free [1] monoid, $A$ a non-empty set on which $W$ acts on the right, and $\cal{A}$ the semicategory with object set $A$ and morphisms triples $(a,w,b)\in A\times (W\backslash\{1\})\times A$ such that $b=a\cdot w$ with composition of morphisms $(a,w,b)\circ(b,v,c)=(a,wv,c)$ . Then every morphism of $\cal{A}$ is a bimorphism and no morphism is invertible.

Proof. Left as exercise.

Theorem 2. Let $W$ and $A$ be as above. The quotient of the free monoid in the morphisms of $\cal{A}$ by the congruence generated by the composition of morphisms, i.e. $(a,w,b)(b,v,c)\sim(a,wv,c)$, is cancellative and invertible-free.

Proof. Left as an exercise.

Denote the monoid in Theorem 2 as $\cal{A}^\circ$.

Theorem 3. Let $W$ and $A$ be as above. Then there exists a cancellative invertible-free monoid $X$, set $B$ of homographies from $W$ to $X$, and bijection $\varphi\colon A\leftrightarrow B$ that satisfies $\varphi(a\cdot w)=\varphi(a)^w$ for all $a\in A$ and $w\in W$.

Proof. Let $\cal{A}$ be as above and let $X=\cal{A}^\circ$. Let $\varphi_a(w)=(a,w,a\cdot w)$ if $w\not=1$ and $\varphi_a(1)=1$ for all $a$.

Note that if $W=\{1\}$, then $\varphi_a=\varphi_b$ for all $a,b\in A$.

[1] For definitions of these terms, see https://mathoverflow.net/questions/378076/can-every-cancellative-invertible-free-monoid-be-embedded-in-a-group and https://mathoverflow.net/questions/379782/is-every-invertible-free-cancellative-monoid-action-represented-by-shifting-ce