Let $X$ be a Banach space. $A$ and $B$ are linear closed and densely defined operators and $\lambda\in\rho(A)\cap\rho(B)$ such that $(\lambda - A)^{-1}-(\lambda - B)^{-1}$ is a Frehholm perturbation then it is well known that $\sigma_e(A) = \sigma_e(B)$ where
$\sigma_{e}(A): = \{\lambda \in\mathbb{C}\,\,\hbox{such that} \ \lambda-A \,\,\hbox{isn't a Fredholm operator on}\, X \}$.
In the proof of this result, the authors suppose $\lambda = 0$? why?(see page 288) paper
That is simply the fact that you can replace $A$ and $B$ with $\lambda-A$ and $\lambda-B$ respectively.
This works because $\sigma(A+\lambda)=\sigma(A)+\lambda$. Indeed, $(A+\lambda)-\gamma$ is invertible if and only if $A-(\gamma-\lambda)$ is invertible (they are the same operator). So $\gamma\in\sigma(A+\lambda)$ if and only if $\gamma-\lambda\in\sigma(A)$, i.e. $\gamma\in\sigma(A)+\lambda$.