Let $R$ be a non-commutative ring. If we have a free left $R$-module $F$, we can consider a basis $\{e_i\}_{i\in I}$ in $F$ and endow $F$ with a $(R,R)$-bimodule structure in the following way: For a generic element $x=\sum_{i\in I}r_ie_i$ in $F$, where $r_i\in R$ are almost all equal to $0$, and for $r\in R$, define $$ x\cdot r=\sum_{i\in I}(r_ir)e_i. $$ More generally, if $f:S\to R$ is a ring homomorphism, one could endow $F$ with a $(R,S)$-bimodule structure defining $$ x\cdot s=\sum_{i\in I}(r_if(s))e_i. $$ My question is: does this $(R,S)$-bimodule structure depend on the choice of basis in $F$? My guess is that it does, since if we have another basis $\{\tilde{e}_j\}_{j\in J}$ in $F$, then we could write $e_i=\sum_{j\in J}r_{ji}\tilde{e}_j$ and thus, if we were to multiply $e_i$ in the right by some scalar $r\in R$ and then pass to the "tilde" basis, we would get $\sum_{j\in J}rr_{ji}\tilde{e}_j$, but if we were to first pass from $e_i$ to the "tilde" basis and then apply our definition of multiplication by $r$ on the right in the "tilde" basis, we would end up with $\sum_{j\in J}r_{ji}r\tilde{e}_j$. Hence, I think that this would show the basis dependence on the bimodule structure definition.
Is my argument right?