I want to check whether the following transformation is correct:
$$ \sum_{s=1}^\infty (1-x)^s \exp(-\lambda)\frac{\lambda^s}{s!} = \exp(-\lambda)\sum_{s=1}^\infty \frac{X^s}{s!}\\ = \exp(-\lambda)\exp(X) = \exp(1-x) $$
Where the first line comes from the Taylor expansion. I'm curious whether that can be applied here, and is the normal way to proceed - and whether there are "but care, this only holds for certain values of $x$" caution tales.
You can certainly bring the constant $e^{-\lambda}$ outside of the sum. The remaining sum is $e^{\lambda(1-x)}-1$. The "$-1$" appears here because the sum as written starts at index $s=1$ rather than $s=0$, so the constant term needs to be accounted for.
Assuming that was not a typo, the result would be $e^{-\lambda x}-e^{-\lambda}$.
Addendum: Also, this is valid for all values of $x$ since this particular sum has an infinite radius of convergence.