Going through some notes from the past quarter, I stumbled across the following equation: \begin{equation} (\tilde{\lambda}I-A)^{-1}=Q(\tilde{\lambda}I-\Lambda)^{-1}Q^* \end{equation} I'm sure it made sense to me at the time, but now I can't get back to the intermediate steps other than the usual eigenvalue decomposition $A=Q\Lambda Q^*$.
$A$ is a normal matrix, $\tilde{\lambda}$ is an eigenvalue of $A+\delta A$, $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ on its diagonal, and $Q$ is unitary.
If we take the inverse of both sides, we see that your equation is equivalent to $$ (\lambda I - A) = Q(\lambda I - \Lambda)Q^* $$ (where I have removed the $\sim$ from your $\lambda$). To see that this is true, it suffices to multiply out the right side of the equation: $$ Q(\lambda I - \Lambda)Q^* = Q(\lambda I)Q^* - Q\Lambda Q^* = \lambda I - A $$ which is what we wanted.