This source states that the unit step function in the Z-domain is $\frac{z}{z-1}$. However, in its derivation it states $\sum_{k=0}^{\infty} z^{-k} = \frac{z}{z-1}$.
But doesn't that last relation only hold true for $z>1$? I don't see how that condition is met. I know that the poles have to be in the unit circle for the system to be stable, but I don't see if and how that connects to the condition above.
What am I missing?
The $z$-transform like the $s$-transform is a tool, and it is important to know what you are analyzing with the tool. Knowing that the output of the tool is $z/(z-1)$ is not sufficient to make conclusions.
In one scenario, you could impulse a linear system and find out that it is the unit step, and you take the z-transform of it, it comes out to be $z/(z-1)$. This is the transfer function of the system
$$ Y(z)/U(z) = z/(z-1), $$ and now you can conclude that this linear system is unstable.
In another scenario, you could be handed a linear system $$ Y(z)/U(z) = 0.5z/(z-0.2), $$ and be asked to find the response to a unit step. So what do you do? You have to first obtain $U(z) = z/(z-1)$, and then substitute it in and so on. Now in this case, it doesn't make sense to say that $U(z)$ is not valid because it is not stable.