Since the mean of $\operatorname{Poi}(\mu)$ is $\mu$, we have
$$ \mathbf{E}[X_{n} \mid X_1,\ldots,X_{n-1}] = X_{n-1} $$
and hence $(X_n)$ is a non-negative martingale. So by the martingale convergence theorem, $(X_n)$ converges a.s. Then by the generalized Borel–Cantelli lemma,
\begin{align*}
\mathbf{P}(X_n = 0 \text{ i.o.})
&= \mathbf{P}\Biggl( \sum_{n=1}^{\infty} \mathbf{P}(X_n = 0 \mid X_1,\ldots,X_{n-1}) = \infty \Biggr) \\
&= \mathbf{P}\Biggl( \sum_{n=1}^{\infty} e^{-X_{n-1}} = \infty \Biggr) \\
&= 1,
\end{align*}
where the last line follows from the fact that $X_n$ converges a.s. Together this and the observation that $X_{n+k} = 0$ for all $k \geq 0$ whenever $X_n = 0$ (this is because $\operatorname{Poi}(0) = \delta_0$), we find that $(X_n)$ is eventually constant with the value $0$ almost surely. Therefore
$$ \lim_{n\to\infty} X_n = 0 \quad \text{a.s.} $$
This then shows that $(X_n)_{n\geq 1}$ cannot converge in $L^1$ or $L^2$.
Alternatively, given that a.s.-convergence of $(X_n)$ has been established, we may directly prove that $X_{\infty} := \lim X_n = 0$ a.s. Indeed, by the law of iterated expectation, for each $s \geq 0$ we have
$$ \mathbf{E}[e^{-sX_n}]
= \mathbf{E}[\mathbf{E}[e^{-sX_{n}}\mid X_{n-1}]]
= \mathbf{E}[e^{X_{n-1}(e^{-s}-1)}]
= \mathbf{E}[e^{-f(s)X_{n-1}}], $$
where $ f(s) = 1 - e^{-s} $ and the moment generating function formula for the Poisson distribution is utilized in the second step. From this recurrence formula, we get
$$ \mathbf{E}[e^{-sX_n}] = e^{-f^{\circ n}(s)\lambda}. $$
However, by noting that $0 < f(s) < s$ whenever $s > 0$, it is not hard to check that $f^{\circ n}(s) \to 0$ as $n \to \infty$ for any $s \geq 0$. By this and the dominated convergence altogether,
$$ \mathbf{E}[e^{-sX_{\infty}}]
= \mathbf{E}[\lim_{n\to\infty} e^{-sX_n}]
= \lim_{n\to\infty} \mathbf{E}[e^{-sX_n}]
= 1.$$
This implies that $e^{-s X_{\infty}} = 1$ a.s. and hence $X_{\infty} = 0$ a.s. as desired.
