I'm staring at
$$ \sum_{b=2}^\infty\sum_{x=2}^b Poisson(b, \lambda) \frac{binomial(x, b, \mu)}{1 - binomial(0, b, \mu)}\frac{1}{x}$$
I can do the two following transformations
$$ \sum_{x=2}^\infty \sum_{b=x}^\infty \frac{e^{-\lambda}\lambda^b}{b!} {b \choose x}\frac{\mu^x (1 - \mu)^{b - x}}{1 - (1 - \mu)^b}\frac{1}{x}$$
and now, letting $b=x+k$:
$$ e^{-\lambda}\sum_{x=2}^\infty \lambda^x\sum_{x+k=x}^\infty \lambda^k\frac{1}{(x+k)!} {(x+k) \choose x}\frac{\mu^x (1 - \mu)^k}{1 - (1 - \mu)^{x+k}}\frac{1}{x}\\ e^{-\lambda}\sum_{x=2}^\infty \frac{\lambda^x\mu^x}{x!}\frac{1}{x}\sum_{k=0}^\infty \frac{\lambda^k(1 - \mu)^k}{k!}\frac{1}{1 - (1 - \mu)^{x+k}}$$
If it weren't for that conditional probability at the end, I could now use the definition of the exponential twice to solve this. In this scenario, is there a trick to use?