Is $R\otimes_R M\cong M$ when $R$ not necessarily commutative? Suppose $R$ is not commutative. I am guessing in general that $R\otimes_R M\cong M$ fails to be true. However, are there any cases where this is still true? For instance group rings for a non-commutative group $G$?
 A: If $R$ is any ring and $M$ is a left $R$-module, then there is always an isomorphism of abelian groups $R \otimes_R M \cong M$ given by $r \otimes m \mapsto rm$ and $m \mapsto m \otimes 1$ in the other direction.
In fact, $(r,m) \mapsto rm$ is $R$-balanced, hence induces a homomorphism $R \otimes_R M \to M$ of abelian groups. Clearly, $m \mapsto m \otimes 1$ is a homomorphism in the other direction, and they are inverse, basically since every pure tensor in $R \otimes_R M$ may be written as $r \otimes m = 1r \otimes m = 1 \otimes rm$.
Commutativity is never used. In fact, it is better to learn the tensor product of bimodules in the first place, which is slightly more general but not really more complicated. If $M$ is some $(R,S)$-bimodule and $N$ is some $(S,T)$-bimodule, then $M \otimes_S N$ is an $(R,T)$-bimodule. All the usual properties hold (such as associativity, or neutral objects as above). One can also think of the tensor product over $S$ as a method to "integrate" the $S$-action "away". This also fits to the coend interpretation of the tensor product as
$$M \otimes_S N = \int^{s \in S} M \otimes N.$$
The left action of $R$ on $M$ survives after this integration, also the right action of $T$ on $N$. Many results about modules over commutative rings are actually results about $(R,S)$-bimodules or sometimes $(R,R)$-bimodules. (The formal part of this story even generalizes to so-called profunctors.)
Here is something where commutativity is used: If $M,N$ are $R$-modules where $R$ is a commutative ring (and you will soon regret that I didn't tell you if these are left or right modules), then there is an isomorphism of $R$-modules $M \otimes_R N \cong N \otimes_R M$ mapping $m \otimes n \mapsto n \otimes m$. More generally, one might hope that if $M$ is some $(R,R)$-bimodule and $N$ is some $(R,R)$-bimodule, and $R$ is any ring, we still get an isomorphism $M \otimes_R N \cong N \otimes_R M$. Let's check if $\alpha : M \times N \to N \otimes M,~(m,n) \mapsto n \otimes m$ is $R$-balanced: $\alpha(mr,n)=n \otimes mr$ and $\alpha(m,rn)=rn \otimes m$, but there is no way to identify these things. We know that $nr \otimes m = n \otimes rm$, but $mr$ has to be distinguished from $rm$ in a general $(R,R)$-bimodule. This problem already appears for $M=N=R$. Thus, there is no isomorphism of $R$-bimodules $R \otimes_R R \cong R \otimes_R R$ mapping $a \otimes b \mapsto b \otimes a$, unless $R$ is commutative!
There is a way to correct this failure of symmetry: if $R$ is any ring, $M$ is a right $R$-module and $N$ is a left $R$-module, then we may view $M$ as a left $R^{op}$-module and $N$ as a right $R^{op}$-module, and we have an isomorphism of abelian groups $M \otimes_R N \cong N \otimes_{R^{op}} M$.
A: If $R$ has a unit, this is still true. $R$ is an $(R,R)$-bimodule, $M$ is a left $R$-module, therefore $R \otimes_R M$ is an $R$-left module with the action given by $r \cdot (x \otimes m) = rx \otimes m$. Define $f : M \to R \otimes_R M$ by $f(m) = 1 \otimes m$. Then $f$ is a morphism of $R$-left modules (it's obviously additive):
$$f(r \cdot m) = 1 \otimes (rm) = (1 \cdot r) \otimes m = r \otimes m = r \cdot (1 \otimes m) = r \cdot f(m)$$
Now checking that $f$ is injective and surjective is exactly the same proof as in the commutative case (or you can define the inverse morphism $r \otimes m \mapsto r \cdot m$ and check again that it's a morphism of $R$-left modules).
