The following proof is essentially taken from Wikipedia: Laurent series:
Let $f(z) = \sum_{n=-\infty}^\infty a_n z^n $ be a Laurent series in the annulus $r < |z| < R$ and assume that $f(z) = 0$ for all $z$.
Let $k$ be an integer and choose any $\rho \in (r, R)$. The series converges uniformly on $|z| = \rho$ so that we can change the order of integration and summation in the following calculation:
$$
0 = \frac{1}{2\pi i} \int_{|z| = \rho }\sum_{n=-\infty}^\infty a_n z^{n-k-1} \, dz
= \sum_{n=-\infty}^\infty a_n \frac{1}{2\pi i} \int_{|z| = \rho } z^{n-k-1} \, dz = \sum_{n=-\infty}^\infty a_n \delta_{nk} = a_k \, ,
$$
i.e. all $a_k$ are zero.
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
$$\begin{align}
\frac{e^{1/z^2}}{z-1}&=-\sum_{m=0}^\infty z^{m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{m=0}^\infty z^{m-n}\\\\
&\overbrace{=}^{p=m-n}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=-n}^\infty z^{p}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\left(\sum_{p=-n}^{0} z^{p}+\sum_{p=1}^\infty z^{p}\right)\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=0}^{n} z^{-p}\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{p=0}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}\\\\
&=-e \sum_{p=0}^\infty z^{p}-\sum_{p=1}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}
\end{align}$$
Best Answer
Letting $z = w - 1$, you want the Laurent expansion of $e^{w-1 \over 2 - w}$ on $|w| > 2$. Noting that ${w -1 \over 2 - w} = -1 - {1 \over w - 2}$, you are looking for the Laurent expansion of $e^{-1}e^{-{1 \over w - 2}}$ on $|w| > 2$.
Changing variables once again, this time to $v = {1 \over w}$, you are seeking the Taylor expansion of $e^{-1}e^{v \over 2v - 1} = e^{-{1 \over 2}}e^{1 \over 4v - 2}$ on $|v| < {1 \over 2}$. So $a_0 = e^{-1}$ and you can find $a_1$ and $a_2$ by taking the first two derivatives of $e^{-{1 \over 2}}e^{1 \over 4v - 2}$ at $v = 0$.