You should have expanded $\frac{1}{1+z}$ via $$\frac{1}{1+z}=\frac{1}{2-(1-z)}=\frac{1}{2}\frac{1}{1-\frac{1-z}{2}}=\frac12\sum_{n=0}^\infty\left( \frac{1-z}{2}\right)^n$$ and it converges in $|z-1|<2$. So about $z=1$ your full Laurent series is $$f(z)=\frac{1}{2}\left[-\frac{1}{1-z}-\frac12\sum_{n=0}^\infty \left(\frac{1-z}{2}\right)^n \right]=-\frac{1}{4}\sum_{n=-1}^\infty \left(\frac{1-z}{2}\right)^n. $$

As I mentioned in the comments, the series you derived does not exists in a punctured neighborhood of the point $z=1$ as it only converges in $|z-1|>2$.

You should have expanded $\frac{1}{1+z}$ via $$\frac{1}{1+z}=\frac{1}{2-(1-z)}=\frac{1}{2}\frac{1}{1-\frac{1-z}{2}}=\frac12\sum_{n=0}^\infty\left( \frac{1-z}{2}\right)^n$$ and it converges in $|z-1|<2$. So about $z=1$ your full Laurent series is $$f(z)=\frac{1}{2}\left[-\frac{1}{1-z}-\frac12\sum_{n=0}^\infty \left(\frac{1-z}{2}\right)^n \right]=-\frac{1}{4}\sum_{n=-1}^\infty \left(\frac{1-z}{2}\right)^n. $$