Does the output of this system have a closed form expression?
You figured out the solution by yourself, so this is just about how to show it formally. First, split the sum into two sums:

$$y[n]=\sum_{k=-\infty}^{\infty}(-1)^ku[k]u[n-1-k]+\sum_{k=-\infty}^{\infty}(-1)^ku[k]u[n-2-k]\tag{1}$$

The unit steps give you upper and lower summation indices, but note that the resulting sums are only valid if the upper summation index is not smaller than the lower one:

$$y[n]=u[n-1]\sum_{k=0}^{n-1}(-1)^k+u[n-2]\sum_{k=0}^{n-2}(-1)^k\tag{2}$$

Now note that you can increase the upper summation index by any even number without changing the value of the sum. The second term on the right-hand side of $(2)$ can thus be rewritten as

$$u[n-2]\sum_{k=0}^{n-2}(-1)^k=u[n-1]\sum_{k=0}^{n}(-1)^k\tag{3}$$

In $(3)$ the argument of the unit step can be changed to $n-1$ because the value of the sum equals zero for $n=1$. Using $(3)$, Eq. $(2)$ can be rewritten as

$$\begin{align}y[n]&amp;=u[n-1]\left(\sum_{k=0}^{n-1}(-1)^k+\sum_{k=0}^{n}(-1)^k\right)\\&amp;=u[n-1]\left(2\sum_{k=0}^{n-1}(-1)^k+(-1)^n\right)\\&amp;=u[n-1]\left((1+(-1)^{n+1})+(-1)^n\right)\\&amp;=u[n-1]\tag{4}\end{align}$$