General Form of $S^{-1}A$ - modules I have been trying to show that if a ring $A$ is absolutely flat then so is the localisation $S^{-1}A$ by any multiplicative set. Now while trying to do this, I asked myself the following: Is there a description for the general form of an $S^{-1}A$ - module?
What I am trying to do to prove my original statement  is this: If $A$ is absolutely flat, consider the exact sequence 
$$0 \stackrel{f'}{\longrightarrow} M' \stackrel{f}{\longrightarrow} M$$
of $S^{-1}A$ modules. Then suppose we tensor this with some $S^{-1}A$ module $N$ and consider the map
$$f \otimes 1: M'\otimes_{S^{-1}A } N \longrightarrow M \otimes_{S^{-1}A} N.$$
Now we know that if $N$ is flat as an $S^{-1}A$ - module, then it is flat as an $A$ - module so that the map $f|_A \otimes 1 : M' \otimes_A N \longrightarrow M \otimes_A N$ is injective. We write $f|_A$ for $f$ viewed as an injective $A$ - module homomorphism, rather than an injective $S^{-1}A$ - module homomorphism. Since localisation preserves exactness, we have that the map 
$$S^{-1}(f|_A \otimes 1) : S^{-1}(M' \otimes_A N) \longrightarrow S^{-1}(M \otimes_A N)$$ is injective. If I can produce some kind of $S^{-1}A$ isomorphism between $M' \otimes_{S^{-1}A} N$ and $S^{-1}(M'\otimes_A N)$, I should be done. This is what I am trying to do, which is why I asked for a general description for what an $S^{-1}A$ - module looks like.
Thanks.
 A: Given a commutative ring $A$ and a multiplicative set $S\subset A$ there is an equivalence of categories between:
a) $S^{-1}A$- modules $N$
b) $A$- modules $M$ for which all homotheties $s_M: M\to M: m\mapsto sm$ are bijective.  
To go from a) to b), just restrict multiplication in $N$ by scalars in $S^{-1}A  $ to multiplication by scalars of $A$ via $A\to S^{-1}A$  
To go from b) to a) keep the additive group structure on $M$  and define multiplication by scalars in $S^{-1}A$ by the formula $\frac{a}{s}\cdot m=(s_M)^{-1}(am)$  
Moreover if $M,M'$ are two $S^{-1}A$-modules, the canonical morphism $$q: M\otimes _AM' \to M\otimes _{S^{-1}A} M':m\otimes m'\mapsto m\otimes m' $$ is bijective .   
Edit
In order to address Benjamin's legitimate question in the comments, let me say a few words on the puzzling isomorphism $q$ above .
In $M\otimes _AM'$ we have the equalities $am\otimes m'=m\otimes am'\; $
In $M\otimes _{S^{-1}A}M'$ we have the supplementary equalities $\frac{a}{s}m\otimes m'=m\otimes \frac{a}{s}m'$, so that $q$
 is a quotient map, only surjective a priori.
The  pleasant result is that $q$ is an isomorphism because these supplementary equalities are automatic. Here is the heart of the calculation,  taking place in $M\otimes _AM' $,   
$\frac{a}{s}m\otimes m'=\frac{a}{s}s\mu\otimes m'=a\mu\otimes m'=\mu\otimes am'=\mu\otimes as\mu'=s\mu\otimes a\mu'=m\otimes \frac{a}{s}s\mu'=m\otimes \frac{a}{s}m $   
and, voilà, $\frac{a}{s}$ has jumped over $\otimes_A$ just as it would jump over $\otimes _{S^{-1}A}$ .
In other words,  the supplementary equalities are already valid in $M\otimes_A M'$ and no quotient is necessary: $q$ is an isomorphism.
A: Let me address the original question first.
Suppose $A$ is absolutely flat. Let $B=S^{-1}A$ and let $Q$ be a $B$-module. Let us show that $Q$ is $B$-flat.
Suppose $$\tag{1} 0\to M\to N\to P\to 0$$ is an exact sequence of $B$-modules. WE want to show that $$\tag{2} 0\to Q\otimes_BM\to Q\otimes_BN\to Q\otimes_BP\to 0$$ is exact. 
Now, every $B$-module is an $A$-module, and every homomorphism of $B$-modules is then automatically an homomorhism of $A$-modules. We can then view (1) as a short exact sequence of $A$-modules and $Q$ as an $A$-module. Since $A$ is absolutely flat, $Q$ is flatm and we have an exat sequence $$\tag{3} 0\to Q\otimes_AM\to Q\otimes_AN\to Q\otimes_AP\to 0.$$ 
You can easily check that

if $V$ and $W$ are $B$-modules, then the obvious map $V\otimes_AW\to V\otimes_BW$ is an isomorphism.

and that this isomorphism plays nicely with maps. It follows form this that the short exact sequence (3) is equal to (2), so in particular the latter is exact, as we wanted.
Next, let's deal with your subquestion. I'll use your notation now. The displayed statement above implies that $M'\otimes_{S^{-1}A} N$ and $M'\otimes_A N$ are isomorphic $A$-modules, so in particular $M'\otimes_A N$ is an $S^{-1}A$-module! As a consequence applying the $S^{-1}(-)$ construction to it does not change it.
A: A solution is given in $\S 7.7.2$ of my commutative algebra notes.
It goes like this: let $N$ be a $S^{-1} R$-module.  Then there is an $R$-module $M$
such that $N \cong_{S^{-1} R} S^{-1} M := S^{-1} R \otimes_R M$.  (This is essentially what Georges is arguing for in his answer.  In my notes, I think this fact appears as an exercise.)  By definition of absolutely flat, $M$ is a flat $R$-module, and thus by the local characterization of flatness (which appears earlier in the chapter) so is $S^{-1} M$.
Added: I think the intermediate result used is more important than the fact that it was used to prove, and it is also relevant to the OP's question "Is there a description for the general form of an $S^{-1} A$-module?"
The key notion here is the following: if $\varphi: R \rightarrow S$ is a ring homomorphism, then then an $S$-module $N$ is said to be extended from R (via $\varphi$, although this is usually suppressed) if there is an $R$-module $M$ such that $N \cong_S M \otimes_R S$.
Off the top of my head, there are two general classes of homomorphisms $\varphi$ such that every $S$-module is extended from $R$:
$\bullet$ $\varphi: R \rightarrow S^{-1} R$ is a localization map (discussed above).
$\bullet$ $\varphi: R \rightarrow R/I$ is a quotient map.  (This is easy.)
However there are many important results about particular classes of $S$-modules being extended (or not!) from $R$.  For instance, suppose $R = k$ is a field, so $S$ is a $k$-algebra.  Then an $S$-module is extended from $k$ iff it is free.  In particular, when $S = k[t_1,\ldots,t_n]$, a famous conjecture of Serre can be stated as: every finitely generated projective $S$-module is extended from $k$.  It would be equivalent to say that every finitely generated projective $S$-module is free, but the above formulation is indeed the path taken in all proofs of Serre's Conjecture I have seen, so this is somehow the more insightful way to say it.
