\begin{equation*}
M(0) = e^{\mu \left ( e^0 - 1 \right )} = e^0 = 1.
\end{equation*}
Continuing,
\begin{equation*}
M'(t) = \mu e^{\left(\mu {\left(e^{t} - 1\right)} + t\right)}
\end{equation*}
and therefore
\begin{equation*}
M'(0) = \mu e^{\left(\mu {\left(1 - 1\right)} + 0\right)} = \mu e^0 = \mu.
\end{equation*}
Continuing with the second derivative,
\begin{equation*}
M''(t) = {\left(\mu e^{t} + 1\right)} \mu e^{\left(\mu {\left(e^{t} - 1\right)} + t\right)}
\end{equation*}
and therefore
\begin{equation*}
M''(0) = {\left(\mu + 1\right)} \mu e^{\left(\mu {\left(1 - 1\right)} + 0\right)} = {\left(\mu + 1\right)} \mu e^0 = \mu + \mu^2
\end{equation*}
which is the squared mean plus the variance for the poisson distribution.