$$f(z)=\frac{e^z-1}{z}$$
Taking the limit $$\lim_{z\to 0}(z-0)\frac{e^z-1}{z}=\lim_{z\to 0}z\frac{e^z-1}{z}=\lim_{z\to 0}{e^z-1}=0$$
On the other hand $$\frac{e^z-1}{z}=\frac{\sum_{n=0}^{\infty}\frac{z^n}{n!}-1}{z}=\sum_{n=0}^{\infty}{\frac{z^{n-1}}{n!}-\frac{1}{z}}=z^{-1}+1+\frac{z^2}{2}+\frac{z^3}{3!}+...-\frac{1}{z}=1+\frac{z}{2}+\frac{z^2}{3!}+...+$$
So which Singularity is it?
$ \lim_{z \to 0}f(z)=\lim_{z\to 0}\frac{e^z-1}{z}=1$, hence $f$ an a removable singularity at $0$ (by Riemann).