The obstruction that Bjørn Kjos-Hanssen describes can be made even worse-- it applies to any algorithm (not just von Neumann's trick) and also applies to randomized algorithms (i.e. even if the number of samples is not limited in advance, only the expected number of samples is limited).

Suppose we fix $\epsilon>0$ and we let $Z$ be a Bernoulli random variable with weight $p=\min(\frac{\alpha}{\beta}+\delta,1)$, where $\delta<\epsilon$. It would be nice if we could sample $Z$ after drawing at most $B(\epsilon)$ samples of $X_i$ and $Y_i$, as opposed to $B(\epsilon,\alpha,\beta)$.

Unfortunately, we're out of luck. Suppose that $\alpha=(1/2)\beta>0$. As we let $\beta\rightarrow 0$, then the entropy produced per sample of $X_i$ or $Y_i$ goes to zero. However, the entropy necessary to sample $Z$ is finite (for $\epsilon<1$). So the expected number of samples required to generate even an approximate solution is unbounded.

And here's a tweak on the "uninteresting" approach mentioned by the original poster, specifically step (1). Let $\tau$ be the first time that $Y=1$, i.e. $Y_1=0,Y_2=0,...,Y_{\tau-1}=0,Y_\tau=1$. (Note that $\tau$ is a random variable.) Define $A$ as
$$ A = \sum_{i=1}^\tau X_i $$
Then note that
$$E[A]=\frac{\alpha}{\beta}$$
This gives us an estimator for $\frac{\alpha}{\beta}$ and is "self-calibrating", in the sense that you don't need to specify $n$ (it just falls out of $\tau$).

It might be tempting to try to use $A$ as a sampler directly. For example, suppose we generate $N$ copies of $A$, say $A_1,...,A_N$, and set $S=\sum_i A_i$. Then $E[S/N]=\alpha/\beta$, and we will converge to the mean as $N$ grows. Unfortunately,
the variance of a weight $\alpha/\beta$ Bernoulli random variable is
$$\frac{\alpha\beta-\alpha^2}{\beta^2}$$
whereas the variance of $A$ is
$$var[A]=\frac{\alpha\beta-\alpha^2}{\beta^2} + \frac{2\alpha^2(1-\beta)}{\beta^2}$$
(note that the first term matches, so $var[A]$ is strictly greater than the target variance for $0<\alpha,\beta<1$). So, the distribution of $S$ and a real Bernoulli random variable will not match even asymptotically, only their means.

(In case it's helpful, here's a handy reference for the formulas for expectation and variance of random sums of random variables.)