= Solution
For $0<x<r$, the exit time is at least one. Since $\{\eta\geq n\}=\{\eta>n-1\}\in\mathcal F_{n-1}$, this event is independent of $X_n$. The <Tonelli theorem> gives
$$
\mathbb E|X_\eta|
=\sum_{n\geq1}\mathbb E\bigl[|X_n|\mathbf1_{\{\eta=n\}}\bigr]
\leq\sum_{n\geq1}\mathbb E\bigl[|X_n|\mathbf1_{\{\eta\geq n\}}\bigr]
=\mathbb E|X_1|\sum_{n\geq1}\mathbb P(\eta\geq n).
$$
Therefore the <integrability of a stopped random-walk increment> bound is
$$
\boxed{\mathbb E|X_\eta|\leq\mathbb E|X_1|\,\mathbb E\eta<\infty.}
$$
It is the survival event, not the exit-at-$n$ event, that is independent of the next increment. The selected exit increment need not have the same distribution or mean as $X_1$.
For $r\leq x$, the printed variable $X_\eta=X_0$ is undefined because the increment sequence starts at one. Either restrict this part to $0<x<r$, or make the harmless additional convention $X_0=0$. With that convention the conclusion also holds in the immediate-exit case. \b[The integrability assertion needs this indexing qualification.]
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