First, we can write two series for $\frac1{z-1}$ in the two regions $|z|<1$ and $|z|>1$ as
$$\frac1{z-1}=\begin{cases}
-\sum_{n=0}^\infty z^n&,|z|<1\\\\
\sum_{n=1}^\infty z^{-n}&,|z|>1\tag1
\end{cases}$$
Second, the Laurent series for $e^{1/z^2}$ for $0<|z|$ is given by
$$e^{1/z^2}=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\tag2$$
where $a_n$ the sequence such hat
$$a_n=\begin{cases}
1&,n\,\text{even}\\\\
0&,n\,\text{odd}
\end{cases}$$
Putting $(1)$ and $(2)$ together reveals
$$\frac{e^{1/z^2}}{z-1}=
\begin{cases}
-\sum_{m=0}^\infty z^m \sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,0<|z|<1\tag3\\\\
\sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,1<|z|
\end{cases}
$$
For $|z|>1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written
$$\begin{align}
\frac{e^{1/z^2}}{z-1}&=\sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\
&=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,\sum_{m=1}^\infty z^{-(n+m)}\\\\
&\overbrace{=}^{p=n+m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=n+1}^\infty\,z^{-p}\\\\
&=\sum_{p=1}^\infty\left(\sum_{n=0}^{p-1} \frac{a_n}{(n/2)!}\right)\,z^{-p}
\end{align}$$
For $0<|z|<1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written
$$\begin{align}
\frac{e^{1/z^2}}{z-1}&=-\sum_{m=0}^\infty z^{m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{m=0}^\infty z^{m-n}\\\\
&\overbrace{=}^{p=m-n}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=-n}^\infty z^{p}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\left(\sum_{p=-n}^{0} z^{p}+\sum_{p=1}^\infty z^{p}\right)\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=0}^{n} z^{-p}\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{p=0}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}\\\\
&=-e \sum_{p=0}^\infty z^{p}-\sum_{p=1}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}
\end{align}$$