Closed form solution to recurrence: $g(n)=(k-2)g(n-1)+(k-1)g(n-2)$

The issue is that your attempt to find the generating function doesn't actually make use of the initial conditions $g_0 = 0$, $g_1 = 1$ at all; you need to encode these somehow into the expression to actually get a useful result. (Also, the recurrence doesn't even hold for $g_1$, so we need to distinguish that case at the very least.) In particular, we can write $$ B(x) = 0 \cdot x^0 + 1 \cdot x^1 + \sum_{n=2}^{\infty} ((k-2)g_{n-1} + (k-1)g_{n-2}) x^n. $$ Can you take it from here?

Alternatively, instead of using generating functions, just construct the characteristic polynomial $\lambda^2 - (k-2) \lambda - (k-1) = 0$, which factorises as $(\lambda - (k-1))(\lambda + 1)=0$. So we can write the general form as $$g_n = A\cdot (k-1)^n + B \cdot (-1)^n,$$ from which you can use your initial conditions to finish.