I tried to solve it this way but I want to check if it is correct
$f(z)=\frac{e^z}{z}+e^\frac{1}{z}=\frac{1}{z}\sum_{n=0}^\infty{\frac{z^n}{n!}}+\sum_{n=0}^\infty{\frac{1}{n!z^n}}$
$=\sum_{n=0}^\infty{\frac{z^{n-1}}{n!}+\frac{1}{n!z^n}}=\sum_{n=0}^\infty{\frac{z^{2n-1}+1}{n!z^n}}$
$=1+\frac{1}{z}+\sum_{n=1}^\infty{\frac{z^{2n-1}+1}{n!z^n}}$
No, that is not correct. A Laurent series centered at $0$ is an expression of the form $\sum_{n=-\infty}^\infty a_nz^n$ and what you got is not of that form.
Your computations are correct though. You should be able to deduce from them that$$(\forall z\in\Bbb C\setminus\{0\}):\frac{e^z}z+e^{1/z}=\sum_{n=-\infty}^{-2}\frac1{(-n)!z^n}+\frac2z+2+\sum_{n=1}^\infty\frac{z^n}{(n-1)!}.$$