In an (R,R)-bimodule on which elements act the same from the left or right, ring elements commute up to action

Let $R$ be a ring and let $M$ be an $(R,R)$-bimodule such that $r \cdot m = m \cdot r$ for all $m \in M$ and $r \in R$. Prove that for all $r,s \in R$ and $m \in M$, $rs \cdot m = sr \dot m$. (That is, the assumption that $R$ is commutative in the definition of an $R$-algebra is a rather natural one.)

Let $r,s \in R$ and $m \in M$. Then we have $rs \cdot m = r \cdot (s \cdot m) = r \cdot (m \cdot s)$ $= (m \cdot s) \cdot r$ $= m \cdot sr$ $= sr \cdot m$, as desired.