For simplicity, denote $\sum_{i = 1}^n X_i$ by $T$. By assumption, $T \sim \text{Poisson}(n\theta)$. Thus for any measurable function $g: \mathbb{R}^1 \to \mathbb{R}^1$ such that $E_\theta[g(T)] = 0$, we have
$$\sum_{k = 0}^\infty g(k) e^{-n\theta}\frac{(n\theta)^k}{k!} = 0, \quad \forall \theta > 0,$$
which implies
$$\sum_{k = 0}^\infty c_k\theta^k = 0, \quad \theta \in (0, \infty), \tag{1}$$
where $c_k = g(k)n^k/k!, k = 0, 1, 2, \ldots$.
Define a power series $f$ on $\mathbb{R}$ by $f(x) = \sum_{k = 0}^\infty c_k x^k$, then $(1)$ implies that the convergence interval of $f$ is the entire real line$^\dagger$. Since $\sum c_k x^k = \sum 0 x^k$ on $E = (0, \infty)$ by $(1)$, and $E$ contains a limit point, it follows by Theorem 8.5 in Principles of Mathematical Analysis by Walter Rudin that $c_k = 0, k = 0, 1, 2, \ldots$, whence $g(0) = g(1) = \cdots = 0$. This implies that
\begin{align}
P_\theta[g(T) = 0] = \sum_{k: g(k) = 0}e^{-n\theta}\frac{(n\theta)^k}{k!}
= \sum_{k = 0}^\infty e^{-n\theta}\frac{(n\theta)^k}{k!} = 1,
\end{align}
i.e., $T$ is complete.
$^\dagger$: $(1)$ implies that for each $\theta_0 > 0$, the series $\sum_k c_k\theta_0^k$ converges. Hence there exists $M > 0$ such that $|c_k\theta_0^k| \leq M$ for all $k$. Therefore, for all $x$ such that $|x| < \theta_0$,
\begin{align}
\sum_{k = 0}^\infty \left|c_k x^k\right| =
\sum_{k = 0}^\infty \left|c_k \theta_0^k\right|\left|\frac{x}{\theta_0}\right|^k \leq
M\sum_{k = 0}^\infty \left|\frac{x}{\theta_0}\right|^k < \infty,
\end{align}
which shows that $\sum_k c_kx^k$ converges (absolutely) on $(-\theta_0, \theta_0)$.
[self-study]tag. Also add the source of the exercise. Thank you. – Reviewer $\endgroup$