Localization of bimodules Let $A$ be a commutative ring, let $S\subset A$ be a multiplicative subset, and let $M$ be a symmetric $A$-bimodule.
Consider $M$ as a left $A\otimes A$-module. We can form the localization of $M$ at $S\otimes S$; it is a left $(A\otimes A)_{S\otimes S}\cong A_S\otimes A_S$-module. 
We can also look at $M$ as merely a left $A$-module, and look at the localization $M_S$ which is a left $A_S$-module, then consider $M_S$ as a symmetric $A_S\otimes A_S$-module.
I feel like these two objects should be isomorphic as left $A_S\otimes A_S$-modules, given that we started from a symmetric bimodule $M$. Is this true?
Note that the first one can also be expressed as $(A_S\otimes A_S) \otimes_{A\otimes A} M$, and the second one as $A_S\otimes_A M$. I feel a bit uneasy dealing with tensor products over different bases...
 A: Yes, these modules are isomorphic.  I think this is easier to see by generalizing the setup a little bit, since the fact that you are considering $A\otimes A$ here is not actually particularly relevant.
Suppose $f:B\to C$ is a homomorphism of commutative rings and $M$ is a $C$-module.  Given a multiplicative subset $T\subseteq B$, the localization $C_{f(T)}$ is naturally isomorphic to the tensor product of $B$-algebras $B_T\otimes_B C$ (that is, the localization of $C$ with respect to $T$ as a $B$-algebra).  If $M$ is a $C$-module, we can then turn it into a $B_T$-module in two different ways.  We can first consider $M$ as a $B$-module via $f$, and then localize at $T$ to get $M_T=B_T\otimes_B M$.  Or, we can first localize $M$ at $f(T)$, and then consider $M_{f(T)}$ as a $B_T$-module via the localized version of $f$ which is a map $B_T\to C_{f(T)}$.  These two modules are isomorphic, since $$M_{f(T)}=C_{f(T)}\otimes_C M=(B_T\otimes_B C)\otimes_C M=B_T\otimes_B M.$$
In your case, $B=A\otimes A$, $C=A$, $f:A\otimes A\to A$ is the canonical map, and $T=S\otimes S$.  Note that $f(T)$ is just $S$, so the previous paragraph gives that if you take an $A$-module $M$ and consider it as an $A\otimes A$ module and then localize at $S\otimes S$, that's the same as first localizing at $S$ and then considering $M_S$ as an $(A\otimes A)_{S\otimes S}$-module.

Alternatively, you can just directly verify that your two modules have the same universal property.  That is, you want to show that $M_S$, when considered as a symmetric $A$-bimodule, has the universal property of the localization of $M$ at $S\otimes S$.  That is, writing $f:M\to M_S$ for the canonical map, you want to show that if $g:M\to N$ is any homomorphism of $A\otimes A$-modules such that $S\otimes S$ acts invertibly on $N$, then there is a unique $A\otimes A$-module map $h:M_S\to N$ such that $g=hf$.
To prove this, let $K\subseteq N$ be the submodule of symmetric elements: that is, the set of $n\in N$ such that $(a\otimes1)n=(1\otimes a)n$ for all $a\in A$.  Note that $K$ is an $(A\otimes A)_{S\otimes S}$-submodule of $N$, since it is just the set of elements annihilated by $a\otimes 1-1\otimes a$ for all $a\in A$.  Since $M$ is symmetric, $f(M)$ is contained in $K$.  Moreover, for any $h:M_S\to N$, $h(M_S)$ is contained in $K$.  So we may assume that $N=K$; that is, that $N$ is symmetric.
Now since all our modules $M$, $M_S$, and $N$ are symmetric, we may equivalently think of them all as just $A$-modules (a bimodule homomorphism between symmetric bimodules is the same thing as just an $A$-module homomorphism).  The existence of a unique $h$ now just comes from the universal property of $M_S$.
