Let $A$ be an $n \times n$ matrix with all nonnegative entries and row sums strictly less than one, let $V$ be an $n \times n$ nonnegative diagonal matrix satisfying $V \leq I$ (entrywise), let $B\equiv\left(I-AV\right)^{-1}$ and $B^{*}\equiv\left(I-A\right)^{-1}$, let $X$ be a vector in the $n$-dimensional simplex (i.e., $x_j \geq 0,\sum_j^n x_j=1$), let $D_1$ and $D_2$ be two strictly positive diagonal $n \times n$ matrices, and finally let $$\tilde{M}\equiv\left(\mathrm{diag}\left\{ B^{T}X\right\} \right)^{-1}B^{T}\left[V\mathrm{diag}\left\{ X\right\} +\left(I-V\right)\mathrm{diag}\left\{ B^{T}X\right\} \right]D_{1}BD_2\mathrm{diag}\left(\iota-A\iota\right).$$ I want to show that the spectral radius of $\tilde{M}$ is less than or equal to one, $\rho(\tilde{M})\leq 1$, provided that the following condition holds $$\tag{1} D_{1} B^{*} D_{2} \left(I-A\right) \iota\leq\iota.$$
It is useful to note the connection between this question and two questions that have already been solved. First, in the special case with $D_2 = I$, the problem above simplifies to showing that $\rho(M)\leq 1,$ where $$M\equiv\left(\mathrm{diag}\left\{ B^{T}X\right\} \right)^{-1}B^{T}\left[V\mathrm{diag}\left\{ X\right\} +\left(I-V\right)\mathrm{diag}\left\{ B^{T}X\right\} \right] B\mathrm{diag}\left(\iota-A\iota\right).$$ This is question Bounding spectral radius of special matrix which has already been solved, see https://math.stackexchange.com/a/4402778/165163. Second, in the special case in which $D_1 = \eta I$ and $D_2 = \gamma \mathrm{diag}(e_i)$, with $\eta,\gamma$ nonnegative constants and $e_i$ the vector with zeros everywhere except for a 1 in position $i$, then $$\rho(\tilde{M}) = \tilde{M}_{ii}=\frac{\sum_{l}\left(v_{l}x_{l}+\left(1-v_{l}\right)\sum_{k}b_{kl}x_{k}\right)b_{li}^{2}\eta\left(1-s_{i}\right)\gamma_{i}}{\sum_{r}b_{ri}x_{r}}.$$ Since both the numerator and numerator are linear in $x$, it is enough to consider this expression at corners, $X=e_j$. Thus, we need to show that $$\tilde{M}_{ii}^{(j)}=v_{j}b_{ji}\eta\left(1-s_{i}\right)\gamma_{i}+\frac{\sum_{l}\left(1-v_{l}\right)b_{jl}b_{li}^{2}\eta\left(1-s_{i}\right)\gamma_{i}}{b_{ji}}\leq1.$$ Given our assumptions on $D_{1}$ and $D_{2}$, condition (1) collapses to $\eta b_{ji}^{*}\left(1-s_{i}\right)\gamma_{i}\leq$ for all $j$. Since $b_{ii}^{*}\geq b_{ji}^{*},\forall j,$ then we would need to show that $$v_{j}b_{ji}^{2}+\sum_{l}\left(1-v_{l}\right)b_{jl}b_{li}^{2}\leq b_{ji} b_{ii}^{*},$$ as formulated in this question Inequality involving matrix inverse elements, which has already been solved (see here). To some extent, the challenge now is to somehow combine the ideas in the solutions to these two special cases to solve the more general problem postulated here.
Finally, as in the case with $D_2 = I$ discussed in Bounding spectral radius of special matrix, two simple cases are illustrative. First, if $V = I$ then condition (1) implies that $\tilde{M}\iota \leq \iota$ and so $\rho(\tilde{M}) \leq 1$. Second, if $A$ is diagonal then $\tilde{M}$ would be diagonal and so we would just need to show that each diagonal element is lower than one. But each of diagonal element of $\tilde{M}$ would be of the form $$d_1d_2\left(v+\frac{1-v}{1-av}\right)\frac{1-a}{1-av},$$ while (1) implies that $d_1d_2\leq 1$, so it is enough to show that $$\left(v+\frac{1-v}{1-av}\right)\frac{1-a}{1-av}\leq 1.$$ Simple algebra shows this to be true.