I think I see some mistake here, you have $$e^{-\lambda}\cdot e^{\lambda(1-p+pe^t)}=e^{-(1-p+pe^t)},$$ while it should be $$e^{-\lambda(p-pe^t)}.$$
Then the first moment will be $\lambda p.$
Using standard probability notation you have probability generating function $G_X(z) = c \cdot \ln(1-\tfrac{z}{2})$ and the corresponding random variable is $X$. From the sums in your question, I will assume that $X$ is a non-negative integer random variable. With this condition, one of the properties of the PGF is that:
$$\mathbb{P}(X=k) = \frac{G_X^{(k)}(0)}{k!}.$$
These probability mass values must sum to one, which imposes a constraint on the PGF, which we can use to find the constant $c$. Now, substituting your specified PGF you get:
$$\begin{equation} \begin{aligned}
G_X^{(k)}(z)
= c \cdot \Big( \frac{d}{dz} \Big)^k \ln(1 - z/2)
&= \begin{cases}
c \cdot \ln(1 - z/2) & & & \text{for } k=0, \\[6pt]
- c \cdot (k-1)! \cdot ( 2-z )^{-k} & & & \text{for } k>0, \\[6pt]
\end{cases} \\[6pt]
\end{aligned} \end{equation}$$
which gives the mass function:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X=k) = \frac{G_X^{(k)}(0)}{k!}
&= \begin{cases}
0 & & & \text{for } k=0, \\[6pt]
- (c/k) \cdot 2^{-k} & & & \text{for } k>0. \\[6pt]
\end{cases} \\[6pt]
\end{aligned} \end{equation}$$
The constraint equation therefore reduces to:
$$\begin{equation} \begin{aligned}
1 = \sum_{k=0}^\infty \mathbb{P}(X=k)
= -c \cdot \sum_{k=1}^\infty \frac{1}{k \cdot 2^k}
= -c \cdot \ln (2). \\[6pt]
\end{aligned} \end{equation}$$
From this constraint we have $c=-1 / \ln (2)$ so your PGF is:
$$G_X(z) = - \frac{\ln(1-\tfrac{z}{2})}{\ln(2)} = 1 - \frac{\ln(2-z)}{\ln(2)},$$
and the corresponding probability mass function is:
$$p_X(k) \equiv \mathbb{P}(X=k) = \frac{1}{\ln (2)} \cdot \frac{1}{k \cdot 2^k} \quad \quad \quad \text{for all } k \in \mathbb{N}.$$
This is the logarithmic series distribution with probability parameter $p=\tfrac{1}{2}$. The CDF, moments, and other properties, can be derived from the mass function.
Best Answer
$ 1=\sum\limits_{r=1}^{\infty} P(Y=r)=k\sum\limits_{r=1}^{\infty} P(X=r)=k(1-P(X=0))=k(1-e^{-\lambda})$. So $k=\frac 1 {1-e^{-\lambda}}$.