Q:

Moment generating function (MGF) of the ratio distribution $\displaystyle\frac{X}{Y}$

If we know the moment generating functions (MGFs) of the random variables $X$ and $Y$ to be $M_{X}(s)$ and $M_{Y}(s)$, respectively.  The MGF of the sum $X+Y$ will $M_{X}(s) \cdot M_{Y}(s)$.
So what is the MGF of the ratio distribution $\displaystyle\frac{X}{Y}$? 

A:

It is nowhere near as neat. 
$$\begin{align}\mathsf M_{X/\, Y}(t) =&~ \mathsf E(e^{tX/\, Y}) \\[1ex]=&~\mathsf E(\mathsf E(e^{tX /\, Y}\mid Y)) \\[1ex]=&~ \mathsf E(\mathsf M_{X\mid Y}(t/Y))\end{align}$$
Now, if $X$ and $Y$ are independent (as required for the summation formula you gave) then this becomes:
$$\begin{align}\mathsf M_{X/\, Y}(t) =&~ \mathsf E(\mathsf M_X(t/Y))
\end{align}$$
But it won't get any simpler than that except perhaps for special cases.