As Dylan suggested, we can use the universal property of the tensor product to show that $R \otimes_R M \cong M$. Now this means we would like to show that the $R$ - module $M$ is equipped with a bilinear map $\pi : R \times M \rightarrow M$ such that for any bilinear map $B : M \times N \rightarrow P$, where $P$ is some other $R$ - module, there exists a unique linear map $L : M \rightarrow P$ such that
$B = L \circ \pi.$
This shouldn't be too hard. Say we are given an $R$ - bilinear map $B : M \times N \rightarrow P$. We can define the map $\pi : R \times M \rightarrow M$ by mapping the pair $(r,m)$ to $rm$. One easily checks that the map $\pi$ is well-defined and bilinear. Now define the map $L : M \rightarrow P$ on "elementary elements" by
$L(m) = B(1,m)$
and extend it additively. I said "elementary elements" because usually we define maps on elementary tensors - but there are none here so I just coined this term. Now your map $L$ is easily seen to be well-defined. For linearity, the fact that it is additive comes from definition of how we defined $L$. Let's check that it is compatible with scalar multiplication: For any $r \in R$, we have that
$\begin{eqnarray*} L(rm) &=& B(1,rm) \\ &=& rB(1,m) \hspace{5mm} \text{(By definition of bilinearity)} \\ &=& r L(m) \end{eqnarray*} $
completing the claim that $L$ was $R$ - linear. Now it remains to check that our map $B : R \times M \rightarrow P$ factors uniquely through the tensor product. The question whether $B$ factors through $M$ is equivalent to asking if first sending $(r,m)$ to $B(r,m)$ in $P$ is equal to first sending $(r,m) \mapsto rm$ in $M$, and then sending $rm$ in $M$ to $P$. But this is clear because
$\begin{eqnarray*} L(rm) &=& B(1,rm) \\ &=& rB(1,m) \\ &=& B(r,m). \end{eqnarray*}$
It is possible to do manipulations like that because $M$ is an $R$ module and the other guy in the direct product is $R$ itself.
It now remains to see why given our maps $B$ and $\pi$, there is a unique $R$ - linear map $L : M \longrightarrow P$. If you have any linear map $L$ out of $M$, in order for an appropriate diagram in question to commute we must have that $L(m) = L(\pi(1,m)) = B(1,m)$. There really is no choice for what $L$ is because it is defined by $B$. Hence there is only one $L$ in question for a given bilinear map $B: R \times M \longrightarrow P$ and uniqueness is proven.
We have shown that the $R$ - module $M$ satisfies the universal property of the tensor product $R \otimes_R M$, and hence must be isomorphic to $M$.
$\hspace{6in} \square$