$\lambda-z-e^{-z}=0$ has one solution in the right half plane
A hint.
If $\operatorname{Re} z > 0$ and $\lambda - z - e^{-z} = 0$ then
$$ |\lambda - z| = e^{-\operatorname{Re} z} < 1. $$
In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-\lambda|<1$.