Let us look at the simple case where $\gamma=\sigma^2=B=\mu=0$ and $A=Z_0=1$. By definition of the Lévy measure $\nu$, for every real number $\theta$,
$$
\mathbb E(\mathrm e^{\mathrm i\theta X_t})=\exp\left(t\int_{\mathbb R\setminus\{0\}}(\mathrm e^{\mathrm i\theta x}-1-\mathrm i\theta x\mathbf 1_{|x|\lt1})\,\nu(\mathrm dx)\right),
$$
thus,
$$
\mathbb E(\mathrm e^{\mathrm i\theta X_t})=A(\mathrm i\theta),\qquad A(z)=\exp\left(t\sum_{n\geqslant1}p^n(\mathrm e^{-nz}-1)\right).
$$
The function $A$ is analytical on the disk $D=\{z\in\mathbb C\mid |z|\lt-\log p\}$ hence there exists some complex sequence $(A_n)_n$ such that, for every $z$ in $D$,
$$
A(z)=\sum_{n\geqslant0}A_n\frac{z^n}{n!}.
$$
Differentiating $2n$ times the identity $\mathbb E(\mathrm e^{\mathrm i\theta X_t})=A(\mathrm i\theta)$ with respect to $\theta$ at $\theta=0$ yields $\mathbb E(X_t^{2n})=A^{(2n)}(0)=A_{2n}$. This proves that, for every $z$ in $D$, $\mathbb E(\cosh(zX_t))$ converges, hence $z\mapsto\mathbb E(\mathrm e^{zX_t})$ is analytic on $D$, and equal to $A$ there. Thus, for every real number $\theta$ such that $|\theta|\lt-\log p$,
$$
\mathbb E(\mathrm e^{\theta X_t})=\exp\left(t\sum_{n\geqslant1}p^n(\mathrm e^{-\theta n}-1)\right).
$$
For example $\mathrm e^{X_t}$ and $\mathrm e^{-X_t}$ are integrable if $p\lt\mathrm e^{-1}$. What happens if $p\geqslant\mathrm e^{-1}$ is that the jumps of length $-n$ for $n\geqslant1$ generated by the discrete part of $\nu$ are too numerous hence $X_t=-\infty$ almost surely, for every $t\gt0$.
The proof when one includes the continuous part of $\nu$ on $(0,+\infty)$ is similar but the final result might depend on a balance between the parameters $p$ and $\lambda$ which describe the behaviour of $\nu((-\infty,-x))$ and $\nu((x,+\infty))$ when $x\to+\infty$ since $\nu((-\infty,-x))=p^{x+o(x)}$ and $\nu((x,+\infty))=\mathrm e^{-\lambda x+o(x)}$.
