You can at first determine the Laurent series for $ e^{z} = \sum_{m=0}^\infty \frac{z^m}{m!}$ and $ \frac1{z-1} = \sum_{n=0}^\infty z^{-(n+1)} , \qquad |z|>1$ independently.
To multiply two Laurent series, we can use this formula or simply calculate $\begin{align}\frac{e^{z}}{z-1}= &\left(\sum_{m=0}^\infty \frac{z^m}{m!} \right) \left(\sum_{n=0}^\infty z^{-(n+1)}\right)\\ &=\sum_{m,n\in\mathbb{Z}}[0\leq m][0\leq n] \frac{z^{m-n-1}}{m!}\\ &=\sum_{k,m} [0\leq m][0\leq m-k-1] \frac{z^{k}}{m!}\\ &= \sum_{k=-\infty}^{-1} z^k \underbrace{\sum_{m=0}^\infty \frac1{m!}}_{e} + \sum_{k=0}^\infty z^k \sum_{m=1+k}^\infty \frac1{m!}. \end{align}$ with $k=m-n-1$ and where I used Iverson's bracket.
Edit: As Robert Israel pointed out, we can still express the last sum using the incomplete $\gamma$-function $\gamma(n,x) = \int_0^x t^{n-1} e^{-t} dt = x^n (n-1)! e^{-x} \sum_{k=0}^\infty \frac{x^k}{(n+k)!} .$ Setting $x=1$, we have with $n\geq 1$ $\gamma(n,1) = (n-1)! \sum_{i=n}^\infty \frac{1}{i!}. $ Thus, $\frac{e^z}{z-1} = e \sum_{k=-\infty}^{-1} z^k+ \sum_{k=0}^\infty\frac{\gamma(k+1,1)}{k!} z^k.$