Well, your proof is okay. Let me suggest a slightly different way of looking at it:
Consider a sequence
$$ 0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C$$
and look at
$$ 0\; \xrightarrow{\phantom{j_{\ast}}}
\operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\;
\operatorname{Hom}{(M,C)},$$
where I write $i_{\ast} = \operatorname{Hom}(M,i)$ and $j_{\ast} = \operatorname{Hom}(M,j)$.
Saying that the first sequence is exact amounts to saying $i = \ker{j}$, that is $ji = 0$ and $i$ has the universal property as depicted in the first diagram below: If $g: M \to B$ is such that $jg = 0$ then there exists a unique $f: M \to A$ such that $if = g$. In other words if $j_\ast g = 0$ then $g = i_{\ast}f$, or yet again $\operatorname{Ker}{j_\ast} \subset \operatorname{Im}{i_{\ast}}$ and $i_{\ast}$ is injective.
On the other hand, the second diagram says: if $g: M \to B$ is of the form $g = if = i_{\ast}f$ then $j_{\ast}g = 0$ (because $j_\ast g = j_{\ast}i_{\ast} f = (ji)_{\ast}f = 0f = 0$). In other words, $\operatorname{Im}{i_{\ast}} \subset \operatorname{Ker}{j_{\ast}}$.
Summing up, we have shown that for all $M$ the sequence
$$ 0\; \xrightarrow{\phantom{j_{\ast}}}
\operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\;
\operatorname{Hom}{(M,C)}$$
is exact both at $\operatorname{Hom}{(M,A)}$ ($i_{\ast}$ is injective) and at $\operatorname{Hom}{(M,B)}$ ($\operatorname{Im}{i_{\ast}} = \operatorname{Ker}{j_{\ast}}$)—you seem to have forgotten about the first point here.
Added: As witnessed by the argument above, left exactness of $\operatorname{Hom}$ is essentially the definition of left exactness in the abelian category of $R$-modules. As the comments try to point out, the importance of this fact cannot be overemphasized.
I would like to add two further points:
A functor $F$ is left exact in your definition if and only if $0 \to F(A) \to F(B) \to F(C)$ is left exact for every short exact sequence $0 \to A \to B \to C \to 0$.
Indeed, in a left exact sequence $0 \to A \to B \to C$, we may factor $j: B \to C$ over its image as $B \twoheadrightarrow \operatorname{Im}{j} \rightarrowtail C$ and obtain two exact sequences $$0 \to A \to B \to \operatorname{Im}{j} \to 0 \qquad \text{and} \qquad 0 \to \operatorname{Im}{j} \to C \to \operatorname{Coker}{j} \to 0.$$ Applying $F$ to these two exact sequences, we obtain the left exact sequences $$0 \to F(A) \to F(B) \to F(\operatorname{Im}{j}) \qquad \text{and} \qquad 0 \to F(\operatorname{Im}{j}) \to F(C ) \to F(\operatorname{Coker}{j}).$$ Since the kernel of a map is not changed by postcomposing the map with a monomorphism (check this!), we have $$\operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}))} = \operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}) \to F(C))},$$ so by functoriality of $F$ we get a left exact sequence $0 \to F(A) \to F(B) \to F(C)$ as desired.
A natural question is: When does $\operatorname{Hom}(M,-)$ send short exact sequences to short exact sequences? In other words, when is $j_\ast = \operatorname{Hom}{(M,j)}$ an epimorphism for all short exact sequences $0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C \to 0$?
In view of left exactness of $\operatorname{Hom}{(M,-)}$ the question is: Given any morphism $h: M \to C$ and any epimorphism $j: B \to C$, when is $h$ of the form $h = j_\ast g$ for some morphism $g: M \to B$?
As you certainly know, this is precisely the definition of projective modules: $M$ is called projective if and only if $g$ always exists, for all epimorphisms $j: B \twoheadrightarrow C$ and all $h: M \to C$. For emphasis:
A module $M$ is projective if and only if $\operatorname{Hom}{(M,-)}$ is exact, that is: it sends short exact sequences to short exact sequences.
This statement follows from $Hom(-,I)$ being an exact functor iff $I$ is injective. To see why this is true, we will argue in the opposite category. If $I$ is injective in $\mathcal{A}$ for any category $\mathcal{A}$, then $I$ is projective in $\mathcal{A}^{op}$ by the definition of the opposite category. Now, recall that the definition of a projective module says $P$ is projective iff $Hom(P,-)$ is exact.
So we have the short exact sequence $0\rightarrow C^*\rightarrow A^*\rightarrow B^*\rightarrow 0$, which is the same as $0\rightarrow Hom(C,\mathbb{Q}/\mathbb{Z}) \rightarrow Hom(A,\mathbb{Q}/\mathbb{Z})\rightarrow Hom(B,\mathbb{Q}/\mathbb{Z}) \rightarrow 0$. Going to the opposite category, we see we get $0\leftarrow Hom(\mathbb{Q}/\mathbb{Z},C) \leftarrow Hom(\mathbb{Q}/\mathbb{Z},A)\leftarrow Hom(\mathbb{Q}/\mathbb{Z},B) \leftarrow 0$. Applying the fact that $\mathbb{Q}/\mathbb{Z}$ is projective in $\mathcal{Ab}^{op}$ and the definition of projective modules, we see that $0\rightarrow B\rightarrow A\rightarrow C$ is exact. (Note that it's important we have that $Hom(A,\mathbb{Q}/\mathbb{Z})=0 \Leftrightarrow A=0$ thus our functor $Hom(-,\mathbb{Q}/\mathbb{Z})$ is faithful.)
Best Answer
It is enough to prove it preserves short exact sequences: $\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and $S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity.
So consider an injective morphism $\varphi\colon M\to N$ and suppose $\;(S^{-1}\varphi)\Bigl(\dfrac ms\Bigr)=0$ in $S^{-1}N$. This means there exists $t\in S$ such that $\;t\mkern1mu\varphi(m)=\varphi(tm)=0$. But then $$\frac ms=\frac{tm}{ts}=\frac0{ts}=0,$$ which shows $\;S^{-1}\varphi\;$ is injective.