If $A$ is invertible with eigenvalue $\lambda$, then $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$.

$\displaystyle Av = \lambda v$

$\displaystyle v = \lambda A^{-1}v$

$\displaystyle \frac{1}{\lambda} v = A^{-1}v$

For this specific problem, just multiply by $v$ and go from there.

$\displaystyle (2A^{-1}+I)v = 2A^{-1}v + v = \frac{2}{\lambda}v + v = \left(\frac{2}{\lambda}+1 \right)v$

So the eigenvalue for $2A^{-1}+I$ would be $\frac{2}{\lambda}+1$.