An example of a bimodule is any R module M, where S is the ring of R endomorphisms of M, written on the right. By definition, (R*M)*S = R*(M*S). That's what we mean by an R endomorphism. Hence we have a bimodule.
Conversely, a bimodule M is an R module, and if f is an element of S, it rearranges the elements of M; a function from M into itself. Since f commutes with R, and with addition in M, it is a valid M endomorphism. Since M is a right S module, functions in S add and multiply as described above. Thus S is a subring of the endomorphism ring of M.
We can build a ring homomorphism from R into S. Let f be an element of R, and let the corresponding function be R*f, i.e. scaling by f on the right. Verify that this is a module homomorphism. Showing (xr)f = x(rf) is merely restating associativity within R.
Next, show the map is a ring homomorphism from R into S. The function induced by f+g is indeed rf+rg. This is a restatement of the distributive property of R. Also, the function induced by fg is the composition of the function of f with the function of g, namely rfg.
Different elements map 1 to different elements in R, and represent distinct endomorphisms in S. The map is injective, and R embeds in S.
The image of 1 completely determines the endomorphism, and when e(1) = x, right multiplication by x induces that very endomorphism. The map is onto, and R = S.
Don't confuse module endomorphisms with ring endomorphisms. Complex conjugation fixes 1, and swaps i and -i. This is a ring endomorphism that fixes 1, but there is no left R module endomorphism that fixes 1, other than the identity map. It's a different world.