Geometric distribution of a Yule process multiplied by $e^{-\lambda s}$ converges to exponential distribution with parameter 1

Question

If Y is a Yule process with rate $\lambda$ than I know that the generating function is $$F(x,t) = \frac{xe^{-\lambda t}}{1-x+xe^{-\lambda t}}$$.

Now I want to show that $$\lim_{s\to \infty}e^{-\lambda s}Y(s)=\epsilon$$ where $\epsilon$ has a exponential distribution with mean 1.

If I define $Z(s):=e^{-\lambda s}Y(s)$ than the generating function should be $$F(x,s) = \frac{xe^{-\lambda s e^{-\lambda s}}}{1-x+xe^{-\lambda s e^{-\lambda s}}}$$ But as $s\to \infty$ I get $$\frac{x}{1-x+x}=x \neq \frac{1}{1-x}$$ because $$\lim_{s \to \infty} e^{-\lambda s e^{-\lambda s}} = 1$$.

Where is my mistake?