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.
**The answer below is no longer relevant since the OP now says that $X\_1$ and $X\_2$ take on strictly positive integer values and are also independent.**

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If you don't like Byron Schmuland's answer with degenerate random
variables, here is an example with Bernoulli random variables 
showing that the information you have is insufficient to determine
the identical marginal distributions of $X\_1$ and $X\_2$.
More information about the joint distribution is needed.

Let $Y$ have pmf $p\_Y(-1) = p\_Y(+1) = 0.2, p\_Y(0) = 0.6$. $X\_1$ and $X\_2$
thus are Bernoulli random variables. The two joint distributions shown
below are consistent with $Y$. Both give rise to identical marginal 
distributions for $X\_1$ and $X\_2$, but the marginal distributions are
different in the two cases:

$$p(0,1) = p(1,0) = 0.2; p(0,0) = p(1,1) = 0.3; 
\Rightarrow X\_1, X\_2 \sim \text{Bernoulli}(0.5)$$

$$p(0,1) = p(1,0) = 0.2; p(0,0) = 2p(1,1) = 0.4; 
\Rightarrow X\_1, X\_2 \sim \text{Bernoulli}(0.4)$$

In fact, for the given pmf of $Y$, we can choose to make
$X\_1, X\_2 \sim \text{Bernoulli}(p)$ for any $p \in [0.2,0.8]$.