Are isomorphisms as modules always isomorphisms as rings? For example, I'm trying to prove that for any ring $R$ and any multiplicative subsets $U,V$ of $R$, we have
$$U^{1}R\otimes V^{-1}R \simeq U^{-1}(V^{-1}R)$$
as rings. I found that they're isomorphic as $U^{-1}R$-modules. Is an isomorphism between two $U^{-1}R$-modules or $R$-modules that happen to be rings also an isomorphism as rings?
EDIT: Just to be clear, I don't want a proof to what I'm trying to prove, but to whether isomorphisms of modules translate into isomorphisms of rings in any way.
 A: This is verrrrry far from true.  For example, the additive group $\mathbf Z^2$ can be given infinitely many different (non-isomorphic) ring structures. 
For two different squarefree integers $d$ and $d'$, the rings $\mathbf Z[\sqrt{d}]$ and $\mathbf Z[\sqrt{d'}]$ (you can think of these as $\mathbf Z[X]/(X^2-d)$ and $\mathbf Z[Y]/(Y^2-d')$) are both isomorphic to $\mathbf Z^2$ as additive groups ($\mathbf Z$-modules) but they are not isomorphic to each other as rings.
A: No, isomorphisms of modules (which happen to be rings) are not necessarily isomorphisms of rings.
For example, let $R = k$ be a field, and consider the two $k$-algebras $A = k[\epsilon]/(\epsilon^2)$ and $B = k\oplus k.$ $A$ and $B$ are isomorphic as $k$-modules (both are $k$-vector spaces of dimension $2$), but the two are not isomorphic as rings ($A$ has nilpotent elements and $B$ does not). 
In order to upgrade an isomorphism of modules to an isomorphism of rings/algebras, you need to check what happens to the multiplicative structure under the module isomorphism. Indeed, you might write down a map of modules which is an isomorphism, but is not a ring homomorphism at all! For example, for $A$ and $B$ as above, we have the $k$-module isomorphism
\begin{align*}
f : A &\to B \\
a + b\epsilon + (\epsilon^2)&\mapsto (a,b).
\end{align*}
However, this is not a ring homomorphism. On the one hand, we have
\begin{align*}
f((a + b\epsilon + (\epsilon^2))(c + d\epsilon + (\epsilon^2)) &= f(ac + (ad + bc)\epsilon + (\epsilon^2))\\
&= (ac,ad + bc),
\end{align*}
but on the other, we have
\begin{align*}
f(a + b\epsilon + (\epsilon^2))f(c + d\epsilon + (\epsilon^2)) &=(a,b)(c,d)\\
&= (ac,bd).
\end{align*}
