If $A$ has eigenvalue $\lambda_A$ $$B = I - c\frac{I-rA}{I-\bar{r}A}$$
How to derive the eigenvalue $\lambda_B$? $$\lambda_B=1-c\frac{1-r\lambda_A}{1-\bar{r}\lambda_A}$$
where $c, r, \bar{r}$ are constants.
This formula occurs at the derivation of speedup modified Newton iteration in solving DAE. There are no details in the paper for the derivation of this linear algebra piece.
It's just some easy matrix algebra. If $Av=\lambda_A v$ then $(I-rA)v= (1-r\lambda_A)v$ and (multiplying by inverses on both sides) $(I-rA)^{-1}v= (1-r\lambda_A)^{-1}v$, which is valid also if you replace $r$ with $\bar{r}$.
Then you just expand $Bv = (I-c(I-rA)(I-\bar{r}A)^{-1})v=\dots$
[EDIT: wim's answer carries out this computation in more detail.]
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 $(I-\beta A)(I-\alpha A)^{-1}$.
From a quick look at your problem, it seems to me that this should follow by writing $(I - \bar{r} A)^{-1}$ as a geometric series, expanding its product with $I - r A$, and then applying the Spectral Mapping Theorem (which says that if $p$ is a polynomial/power series, then the eigenvalues of $B = p(A)$ are the eigenvalues of $A$ transformed by $p$).
