A semigroup $S$ is moving if $S$ is infinite, and for all finite $F\subseteq S$ and infinite $A\subseteq S$, there are $a_{1},\dots,a_{k}\in A$ such that, for all but finitely many $s\in S$, $$ \{a_{1}s,\dots,a_{k}s\} \nsubseteq F. $$ A function $f\colon X\to Y$ is finite to one if for each $y\in Y$, the set $\{x\in X : f(x)=y\}$ of preimages of $y$ is finite.

The following implications hold among the properties listed below: $$ (1)\Rightarrow(2)\Rightarrow(4)\Rightarrow(5); (1)\Rightarrow(3)\Rightarrow(5). $$ (1) $S$ is a group.

(2) $S$ is left cancellative: for all $a,b,c\in S$, if $ca=cb$ then $a=b$.

(3) $S$ is right cancellative: for all $a,b,c\in S$, if $ac=bc$ then $a=b$.

(4) Left multiplication in $S$ is finite-to-one: for all $a\in S$, the function $x\mapsto ax$ is finite-to-one.

(5) $S$ is moving.

Question. Let $S$ be a smigroup such that right multiplication in $S$ is finite-to-one (that is, for all $a\in S$, the function $x\mapsto xa$ is finite-to-one). Is $S$ necessarily moving?

Remarks: Moving semigroups are interesting since a genrealization of Hindman's Finite Sums coloring theorem applies to them. This is so because a semigroup $S$ is moving if and only if the Stone-Cech remainder $\beta S\setminus S$ is a subsemigroup of $\beta S$.

Theorem. Let $S$ be a moving semigroup. For each coloring of the elements of $S$ in finitely many colors, there are distinct elements $a_{1},a_{2},\dots\in S$ such that all products $a_{i_1}a_{i_2}\cdots a_{i_n}$ with $i_1<i_2<\cdots<i_n$ ($n$ arbitrary) have the same color.

Update: Shevrin's classification of semigroups (On the theory of periodic semigroups, Izvestija Vys\v{s}ih U\v{c}ebnyh Zavedeni\u{\i} Matematika, 1974) implies that, if right multiplication in $S$ is finite-to-one, then $S$ has a moving subsemigroup. It follows that the above coloring theorem holds true for semigroups with finite-to-one right multiplication. But I think the question still makes sense: A positive answer would make the detour through Shevrin's theory unnecessary.