Probability generating function of mixture of discrete random variables 
Consider a discrete distribution $X$ that is a mixture of two
  discrete distributions $A$ and $B$. Explicitly, $X=A$ with probability
  $p$ and $X=B$ with probability $1-p$. Denote the pgfs of $A$ and $B$
  as $\mathcal{G}_A$ and $\mathcal{G}_B$ respectively.

I want to write an expression for the pgf $\mathcal{G}_X$ of $X$.
Here's what I've done so far, am I on the right track?
Let $Y\sim X$ and denote the probability measure on $A$ and $B$ as $\mathbb{P}_A$ and $\mathbb{P}_B$ respectively. Then:
$$
\begin{align*}
\mathbb{P}(Y=k) &= \begin{cases}
\mathbb{P}_A(Y=k) & \textrm{with probability $p$} \\
\mathbb{P}_B(Y=k) & \textrm{with probability $1-p$}
\end{cases} \\
&= p\mathbb{P}_A(Y=k) + (1-p)\mathbb{P}_B(Y=k)
\end{align*}
$$
Then
$$
\begin{align*}
\mathcal{G}_X(z) &= \sum_{k=0}^{\infty}z^k\mathbb{P}(Y=k) \\
&= \sum_{k=0}^{\infty}z^k\left(p\mathbb{P}_A(Y=k) + (1-p)\mathbb{P}_B(Y=k)\right) \\
&= \sum_{k=0}^{\infty}pz^k\mathbb{P}_A(Y=k) + \sum_{k=0}^{\infty}(1-p)z^k\mathbb{P}_B(Y=k) \\
&= p\sum_{k=0}^{\infty}z^k\mathbb{P}_A(Y=k) + (1-p)\sum_{k=0}^{\infty}z^k\mathbb{P}_B(Y=k) \\
&= p\mathcal{G}_A(z)+(1-p)\mathcal{G}_B(z)
\end{align*}
$$
 A: The solution and the general method are both correct, but some of your notation is unnecessary (e.g., there is no need to define $Y$ at all).  Note that this result is just a special case of a more general probability rule, which holds that that the probability generating function of any mixture random variable can be written as a weighted average of the probability generating functions of its mixture parts (with weights equal to the mixture probabilities).

Theorem: Let $X$ be a mixture of the random variables $A_1,...,A_k$ with mixture probabilities $p_1,...,p_k$, then the probability generating function of $X$ 
  can be written as:$$\mathcal{G}_{X}(z) = \sum_{i=1}^k p_i \cdot \mathcal{G}_{A_i}(z).$$
Proof: From the stated mixture form we have $X = A_H$ where $H \sim \text{Categorical}(p_1,...,p_k)$.  Thus, using the law of total expectation you have:$$\begin{equation} \begin{aligned}
\mathcal{G}_{X}(z) 
= \mathbb{E}(z^X) 
&= \mathbb{E}(z^{A_H}) \\[6pt]
&= \mathbb{E}( \mathbb{E}(z^{A_H}|H)) \\[6pt]
&= \mathbb{E}( \mathcal{G}_{A_H}(z)) \\[6pt]
&= \sum_{i=1}^k p_i \cdot \mathcal{G}_{A_i}(z), \\[6pt]
\end{aligned} \end{equation}$$ which was to be shown.  $\blacksquare$

Observe that nothing in this proof requires any assumption about the distributions of the random variables $A_1,...,A_k$ to be discrete.
