It is easy to see that
$$
\exp\biggl(\frac{z}{1-z}\biggr)
=\exp\biggl(\frac{1}{2-(z+1)}-1\biggr)
=\frac{1}{e}\exp\biggl(\frac{1}{2-u}\biggr), \quad u=z+1.
$$
By the Faa di Bruno formula and some properties for the Bell polynomials of the second kind, we obtain
\begin{align*}
\biggl[\exp\biggl(\frac{1}{2-z}\biggr)\biggr]^{(n)}
&=\sum_{k=0}^{n}\exp^{(k)}\biggl(\frac{1}{2-z}\biggr)B_{n,k}\biggl(\frac{1!}{(2-z)^2},\frac{2!}{(2-z)^3},\dotsc, \frac{(n-k+1)!}{(2-z)^{n-k+2}}\biggr)\\
&=\sum_{k=0}^{n}\exp\biggl(\frac{1}{2-z}\biggr) \frac{1}{(2-z)^{n+k}} B_{n,k}(1!,2!,\dotsc,(n-k+1)!)\\
&=\sum_{k=0}^{n}\exp\biggl(\frac{1}{2-z}\biggr) \frac{1}{(2-z)^{n+k}} \binom{n-1}{k-1}\frac{n!}{k!}\\
&\to n!\frac{\operatorname{e}^{1/2}}{2^n}\sum_{k=0}^{n}\binom{n-1}{k-1}\frac{1}{(2k)!!}
\end{align*}
as $z\to0$, where $B_{n,k}(x_1,x_2,\dotsc, x_{n-k+1})$ denotes the Bell polynomials of the second kind. Consequently, we acquire
\begin{equation*}
\exp\biggl(\frac{1}{2-z}\biggr)=\operatorname{e}^{1/2}\sum_{n=0}^{\infty}\Biggl[\frac{1}{2^n}\sum_{k=0}^{n} \binom{n-1}{k-1}\frac{1}{(2k)!!}\Biggr]z^n, \quad z\in[-1,1).
\end{equation*}
Consequently, we obtain
\begin{equation*}
\exp\biggl(\frac{z}{1-z}\biggr)
=\frac{1}{\operatorname{e}^{1/2}}\sum_{n=0}^{\infty}\Biggl[\frac{1}{2^n}\sum_{k=0}^{n} \binom{n-1}{k-1}\frac{1}{(2k)!!}\Biggr](z+1)^n, \quad z\in[-2,0).
\end{equation*}
References
- https://math.stackexchange.com/a/4700985
- https://math.stackexchange.com/a/4253870