Benveniste-Scheinkman condition gives derivative that still depends on the value function
I guess I will provide a basic answer for now, since I somewhat figured this out, and hopefully either someone posts a more complete answer, or I add more to this later.

Basically, we can use forward or backward iteration. For example, for forward iteration (given that we know the value of $V_1$ at $t=0$, call it $V_{initial}$), then we can use this to find the value of $V$ at time $t$ simply by using the equation over and over again. Or we can work backwards, if for some reason forward iteration doesn't work (at this moment I'm not sure why/how this is different then forward). I don't think the solution is too neat, but it works.

Or, if we are at a steady state, then $V_1(K,I_{-1}) = V_1(K'_I)$, so we can simply rearrange the equation and solve.

Again, this isn't a great answer, but it is correct.