I guess you want to compare $M$ with $S^{-1}M$. The answer depends on $S$.
Let $I=\mathrm{Ann}_A(M)$ be the annihilator of $M$.
If $S\cap I\ne\emptyset$, then $S^{-1}M=0$.
More generally, we have $S^{-1}M=\{ x/s \mid x\in M, s\in S\}$. Let $F$ be the set of $x\in M$ such that $S\cap \mathrm{Ann}_A(x)=\emptyset$. Then $x/s=0$ if and only if $x\notin F$. So $S^{-1}M=\{ x/s \mid x\in F, s\in S\}$.
Now suppose $S\cap \mathrm{Ann}_A(x)=\emptyset$ for all $x\in M$. Then the canonical map $M\to S^{-1}M$ is injective. Let us show it is an isomorphism.
Let $x/s\in S^{-1}M$. Consider the multiplication-by-$s$ map $M\to M$, $y\mapsto sy$. It is injective hence bijective because $M$ is finite. So there exists $y\in M$ such that $x=sy$. Thus $x/s=y/1$ belongs to the image of $M\to S^{-1}M$ and we are done.
In your case, to have the same cardinality, you need $p\notin \mathfrak m$ for any maximal ideal of $O_K$ containing $\mathfrak a$.
Edit. Under the finiteness hypothesis on $M$, the canonical map $M\to S^{-1}M$ is always surjective.
Proof: Fix an $x/s\in S^{-1}M$. We have an increasing sequence of submodules $M_n=\{ y\in M \mid s^ny=0\}.$ As $M$ is finite, this sequence is stationary. Suppose $M_{r}=M_n$ for all $n\ge r$. Consider the (well defined) multiplication-by-$s^r$ map on $M/M_r$. If $s^r\bar{y}=0$, then $s^ry\in M_r$ and $s^r(s^ry)=0$. Hence $y\in M_{2r}=M_r$ and $\bar{y}=0$. Therefore $M/M_r\to M/M_r$ is injective, thus bijective. Let $y\in M$ such that $s^ry=x$ in $M/M_r$. We have $s^ry=x+z$ with $s^rz=0$. So $ s^{r-1}y=x/s+ zs^r/s^{r+1}=x/s.$