The answer is no. Here is a monoid where right multiplication is finite-to-one but is not moving.

Let $M$ be the monoid with presentation

$\langle t,x_0,x_1,\ldots,\mid x_0t=x_0,x_it=x_{i-1}, i>0\rangle$.

Then each element of $M$ can be written uniquely in the form $t^nw$ with $n\geq 0$ and $w$ a word over the $x_i$ possibly empty. It is easy to check multiplication on the right is finite-to-one. Each $x_i$ acts injectively on the right. The element $t$ is at most two-to-one: $t^nw$ has 2 preimages under right multiplication by $t$ iff $w$ ends in $x_0$. Compositions of finite-to-one maps are finite-to-one giving the general case.

Take $F=\{x_0\}$ and $A=\{x_0,x_1,\ldots\}$. Then infinitely many powers of $t$ right multiply any finite subset of $A$ into $F$. So $M$ is not moving.

The answer is no. Here is a monoid where right multiplication is finite-to-one but is not moving.

Let $M$ be the monoid with presentation

$\langle t,x_0,x_1,\ldots,\mid x_0t=x_0,x_it=x_{i-1}, i>0\rangle$.

Then each element of $M$ can be written uniquely in the form $t^nw$ with $n\geq 0$ and $w$ a word over the $x_i$ possibly empty. It is easy to check multiplication on the right is finite-to-one. Each $x_i$ acts injectively on the right. The element $t$ is at most two-to-one: $t^nw$ has 2 preimages under right multiplication by $t$ iff $w$ ends in $x_0$. Compositions of finite-to-one maps are finite-to-one giving the general case.

Take $F=\{x_0\}$ and $A=\{x_0,x_1,\ldots\}$. Then infinitely many powers of $t$ right multiply any finite subset of $A$ into $F$. So $M$ is not moving.

Added May 2, 2014. A finitely generated counterexample is the monoid $N$ with presentation

$\langle a,b,t\mid at=a,ab^nt=ab^{n-1}, n>1\rangle$

If you set $x_i=ab^i$, then the $x_i$ and $t$ generate a submonoid isomorphic to $M$ above and hence $N$ is not moving. Since this presentation is Church-Rosser it is easy to check that right multiplication by $a,b$ is injective and right multiplication by $t$ is at most 2-to-1. The normal forms are words of the form $u$ and $uw$ where $u$ is a word in $b,t$ and $w$ is a word in $a,b$ with at least one $a$.

The answer is no. Here is a monoid where right multiplication is finite-to-one but is not moving.

Let $M$ be the monoid with presentation

$\langle t,x_0,x_1,\ldots,\mid x_0t=x_0,x_it=x_{i-1}, i>0\rangle$.

Then each element can be written uniquely in the form $t^nw$ with $n\geq 0$ and $w$ a word over the $x_i$ possibly empty. It is easy to check multiplication on the right is finite-to-one. Each $x_i$ acts injectively on the right. The element $t$ is finite-to-one. The element $t$ is at most two-to-one. Compositions of finite-to-one maps are finite-to-one giving the general case.

Take $F=\{x_0\}$ and $A=\{x_0,x_1,\ldots\}$. Then infinitely many powers of $t$ right multiply any finite subset of $A$ into $F$. So $M$ is not moving.

The answer is no. Here is a monoid where right multiplication is finite-to-one but is not moving.

Let $M$ be the monoid with presentation

$\langle t,x_0,x_1,\ldots,\mid x_0t=x_0,x_it=x_{i-1}, i>0\rangle$.

Then each element of $M$ can be written uniquely in the form $t^nw$ with $n\geq 0$ and $w$ a word over the $x_i$ possibly empty. It is easy to check multiplication on the right is finite-to-one. Each $x_i$ acts injectively on the right. The element $t$ is at most two-to-one: $t^nw$ has 2 preimages under right multiplication by $t$ iff $w$ ends in $x_0$. Compositions of finite-to-one maps are finite-to-one giving the general case.

Take $F=\{x_0\}$ and $A=\{x_0,x_1,\ldots\}$. Then infinitely many powers of $t$ right multiply any finite subset of $A$ into $F$. So $M$ is not moving.