Let $A$ denote matrix based on another matrix $B$: \begin{align} A = (I + \lambda B)^{-1} \end{align} $I$ is the identity matrix and $\lambda$ is a coefficient.
Decomposing $B$ as $USU^T$ where $U^TU=I$: \begin{align} A = (I + \lambda USU^T)^{-1} = U(I + \lambda S)^{-1}U^T \quad \text{(eq. 1)} \end{align} and the trace of $A$ is given as: \begin{align} \operatorname{tr}(A) = \operatorname{tr}\left(U(I + \lambda S)^{-1}U^T\right) = \operatorname{tr}\left((I + \lambda S)^{-1}U^TU\right) = \sum_{i=1}^n \frac{1}{1 + \lambda s_{ii}} \end{align} where $s_{ii}$ are the eigenvalues of $B$.
Given diagonal matrix $X$, we define a new matrix $C$: \begin{align} C = (X + \lambda B)^{-1} X \end{align} Do the identities in eq. (1) hold when there is an arbitrary diagonal matrix in place of the identity matrix such that the trace of $C$ can be written simply as shown below? \begin{align} \operatorname{tr}(C) = \sum_{i=1}^n \frac{x_{ii}}{x_{ii} + \lambda s_{ii}} \end{align}
If $X$ commutes with $B$, and (possibly not necessarily) if $X$ or $B$ is invertible, then the claim is true for $\lambda$ such that $X+\lambda \,B$ is invertible, provided that the ordering of $\left(x_{i,i}\right)_{i=1,2,\ldots,n}$ and the ordering of $\left(s_{i,i}\right)_{i=1,2,\ldots,n}$ are compatible. If not, then the claim can fail. Here is a counterexample.
Let $X:=\begin{bmatrix}1&0\\0&2\end{bmatrix}$ and $B:=\begin{bmatrix}0&1\\1&0\end{bmatrix}$. Then, $x_{1,1}=1$, $x_{2,2}=2$, and $\big\{s_{1,1},s_{2,2}\big\}=\{-1,+1\}$. Now, for $\lambda:=1$, we get $$C=(X+\lambda\,B)^{-1}\,X=\begin{bmatrix}1&1\\1&2\end{bmatrix}^{-1}\,\begin{bmatrix}1&0\\0&2\end{bmatrix}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}\,\begin{bmatrix}1&0\\0&2\end{bmatrix}=\begin{bmatrix}2&-2\\-1&2\end{bmatrix}\,.$$ Thus, $$\text{trace}(C)=4\neq \sum\limits_{i=1}^2\,\frac{x_{i,i}}{x_{i,i}+\lambda\,s_{i,i}}\,,$$ as the only way the right-hand side is defined is when $s_{1,1}=+1$ and $s_{2,2}=-1$, which makes $$ \sum\limits_{i=1}^2\,\frac{x_{i,i}}{x_{i,i}+\lambda\,s_{i,i}}=\frac{1}{1+1}+\frac{2}{2-1}=\frac{5}{2}\,.$$