This comes from N. Jacobson's Basic Algebra 1, 1.4 #1:
Let $A$ be a monoid, $M(A)$ the monoid of transformations of $A$ into itself, $A_L$ the set of left translations, $A_R$ the right translations. Show that $A_R$ is the centralizer of $A_L$ in $M(A)$ and vice-versa. Further show that $A_L\cap A_R$ are translations by elements in the center of $A$
First I'll include what I have, which was absolutely no trouble at all. $(a_L\cdot b_R)$ is the transformation that maps $k \rightarrow akb$. This is the same as the transformation $(b_R\cdot a_L)$. So the left and right translations commute, which means $A_R \subset C(A_L)$ and $A_L \subset C(A_R)$ Similarly, if $c_L=c_R$ then $ca=ac, \forall a\in A$. So $c$ is in the $A$'s center. The other direction is no harder.
Where I am stuck is showing the other direction of the first two inclusions. I tried to start with the assumption that some transformation $j$ was in the centralizer of $A_L$ with the goal to show that $j$ is a right translation.
I got that if there was such a transformation $j$, it would have the following property by virtue of being in $C(A_L)$: $$\forall a, k \in A, j(ak)=a\cdot j(k)$$ Which makes me think that there may be some sweet choice of $a,k$. However, because we are inside a monoid of transformations, I can't come up with any good ideas for picking $a$ and $k$. $j$ doesn't have to be injective, nor surjective, A is not necessarily finite, ect... The only thing we really have in a monoid is an identity, and letting one or both of $a$ and $k$ be $1$ doesn't seem to lead anywhere.
You are almost there. Just take $k = 1$ in your formula to get $j(a) = aj(1)$, which shows that $j = c_R$ with $c = j(1)$.