If you have the adjacency matrix and want to compute the expected time to stay in $V^-$ (for instance) starting from various vertices in $V^-$, then we first want to simplify the adjacency matrix to get rid of the useless information about $V^+$. We want to work with the $|V^-|$ by $|V^-|$ matrix $A$ where the $(u,v)$ entry is still $w_{uv}$, but only the vertices from $V^-$ are present. (So instead of summing to $1$, row $u$ should sum to the probability that a transition from $u$ leads to another vertex in $V^-$.)
Then if we start with a column vector $\mathbf x$ describing a probability distribution over states in $V^-$ (which could just have probability $1$ on a single state), then $A \mathbf x$ will be a sort-of-probability distribution over the state after one step: it will tell us the probabilities for every next state in $V^-$, but transitions to $V^+$ will just disappear. The sum of the components of $A\mathbf x$, which is $\mathbf 1^{\mathsf T}A\mathbf x$ algebraically, will be the probability of staying in $V^-$ after one step.
More generally, $\mathbf 1^{\mathsf T}A^k \mathbf x$ will give us the probability of staying in $V^-$ after $k$ steps. So if we want to know the total number of vertices we visited in $V^-$ before moving to $V^+$, that's going to be $\mathbf 1^{\mathsf T}(I + A + A^2 + \dots )\mathbf x$, which simplifies to $\mathbf 1^{\mathsf T}(I - A)^{-1}\mathbf x$.
So $\mathbf 1^{\mathsf T}(I - A)^{-1}$ tells you the vector of expected lengths of stays in $V^-$ for all starting vertices in $V^-$. (It counts the starting vertex as well, so if a vertex always transitions to $V^+$, it'll have an expected length of $1$.)
We can do this for both $V^+$ and $V^-$ at the same time if we work with the bigger matrix $$A = \begin{bmatrix}A^- & 0 \\ 0 & A^+\end{bmatrix},$$ where $A^-$ is the $V^- \to V^-$ transition matrix and $A^+$ is the $V^+ \to V^+$ transition matrix. (So we take the adjacency matrix for the graph and wipe out the entries that go from one set to the other.) But this is a block matrix and the inverses $(I - A^-)^{-1}$ and $(I - A^+)^{-1}$ are going to be completely independent, anyway.