Let $(\lambda,x)$ be an eigenpair of matrix $A$, so $A x=\lambda x$.
Now we compute $B x$, and see if it is equal to
$\left(1-c\frac{1-r\lambda}{1-\bar{r}\lambda}\right)x$.
$$ \begin{align}
B x & = \left( I - c\frac{I-rA}{I-\bar{r}A} \right) x\\
& = x - c\frac{x-rAx}{I-\bar{r}A} = x - c\frac{x-r \lambda x}{I-\bar{r}A}\\
& = x - c\frac{1-r \lambda }{I-\bar{r}A}x\\
& = x - c\frac{1-r \lambda }{I-\bar{r}A} \, \frac{I-\bar{r}\lambda}{I-\bar{r}\lambda}x\\
& = x - c\frac{1-r \lambda }{I-\bar{r}A} \, \frac{x-\bar{r}Ax}{I-\bar{r}\lambda}\\
& = x - c\frac{1-r \lambda }{I-\bar{r}A} \, \frac{I-\bar{r}A}{I-\bar{r}\lambda}x\\
& = \left( 1 - c\frac{1-r \lambda }{I-\bar{r}\lambda}\right)x\\
\end{align}
$$
Hence, we may conclude that following relationship between the eigenvalues of $A$ and $B$ holds:
$$\lambda_B = 1 - c\frac{1-r \lambda_A }{I-\bar{r}\lambda_A} ,$$
as long as $I-\bar{r}A$ is invertible.
Note that it is not necessary that $A$ is diagonalizable.
In the comments there has been some discussion whether or not $(I-\alpha A)^{-1}$ commutes with $I-\beta A$. If we assume that $I-\alpha A$ is invertible, then
$$ \begin{align}
(I-\alpha A)^{-1}(I-\beta A)
& = (I-\alpha A)^{-1}(I-\beta A)(I-\alpha A)(I-\alpha A)^{-1}\\
& = (I-\alpha A)^{-1}(I-\beta A -\alpha A +\alpha \beta A^2)(I-\alpha A)^{-1}\\
& = (I-\alpha A)^{-1}(I-\alpha A)(I-\beta A)(I-\alpha A)^{-1}\\
& = (I-\beta A)(I-\alpha A)^{-1} ,\\
\end{align}
$$
which shows that indeed $(I-\alpha A)^{-1}$ commutes with $I-\beta A$. Therefore
we can write $\frac{I-\beta A}{I-\alpha A}$ instead of $(I-\alpha A)^{-1}(I-\beta A)$ or $(I-\beta A)(I-\alpha A)^{-1}$.