Question: Ahlfors says that once the existence of the quotient $\frac{a}{b}$ has been proven, its value can be found by calculating $\frac{a}{b} \cdot \frac{\bar b}{\bar b}$. Why doesn't this manipulation show the existence of the quotient?

$\frac{a}{b} = \frac{a}{b} \cdot \frac{\bar b}{\bar b} = a\bar b \cdot b^{-1} \cdot\bar b^{-1} = a\bar b \cdot (b\bar b)^{-1}$, the last term clearly exists since the thing being inversed is real.

Answer: 
I will consider the case when $X\_1$ and $X\_2$ are *independent identically distributed* and integer-valued.

The following properites hold:

1. Since $Y$ is discrete with finite support, its pmf is a polynomial in $z$ and $z^{-1}$ with non-negative coefficients.
$$
 Z\_Y(z) = \sum\_{k=m(Y)}^{n(Y)} \mathbb{P}(Y=k) z^k
$$
2. Let $X$ be the purported solution with $Z\_X(z) = \sum\_{k=m(X)}^{n(X)} \mathbb{P}(X=k) z^k$.
3. Since $Y = X\_1-X\_2$, for $X\_1$, $X\_2$ iids, $Z\_Y(z) = Z\_{X}(z) Z\_{X}(z^{-1})$.

The problem you are posing is whether, given $Z\_Y(z)$ one can determine $Z\_X(z)$ such that $Z\_Y(z) = Z\_X(z) Z\_X(z^{-1})$. Observe that the solution, if exists, is not unique, since $Z\_{X^\prime}(z) = z^k Z\_{X}(z)$ would also be a solution for arbitrary $k \in \mathbb{Z}$. 

The necessary condition for existence of a solution is that $Z\_Y(z)$ should verify $Z\_{Y}(z) = Z\_Y(z^{-1})$. Assuming that the given $Z\_Y(z)$ satisfies this property, finding $Z\_X(z)$ reduces to polynomial factorization.
No. If $X\_1$ and $X\_2$ are identically equal to any constant, then $Y$ is identically equal to zero.