We first have $$ \frac{1}{z-1} = \frac{1}{z}\frac{1}{1-1/z} = \sum_{n=-\infty}^{-1} z^n.$$ We also have $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n.$$ Now here is the perceived issue. Since $f(z)$ is equal to 0, it seems that we have two Laurent series expansions of the same functions. This is contradictory to the property that the Laurent series expansion should be unique.
What am I missing?
The first converges for $|1/z| < 1$ and the second converges for $|z|< 1$. No contradiction.