Suppose $X_n$ is a Markov chain in $S$, then for $t > 0$ the function $h(x)=\E^x e^{tH_A}$ (provided it exists) solves for all $x\notin A$ the equation: $Ph(x)=e^{-t}h(x)$. 2. If $X_n$ is the simple random walk on $\Z$, $A=\{x\in\Z:|x|\geq p\}$, then for all $|x| < p$: $h(x+1)+h(x-1)=2e^{-t}h(x)$ and $h(\pm p)=1$. 3. Put $\cos\vp=e^{-t}$ and $\sin\vp=(1-e^{-2t})^{1/2}$, then
$$
\forall |x| < p:\quad
h(x)=\frac{\cos(\vp x)}{\cos(\vp p)}
$$
provided that $\vp p < \pi/2$, i.e. $t < -\log\cos(\pi/2p)$.
On the set $[H_A\geq1]$ we have $H_A=H_A\circ\Theta+1$ and thus
\begin{eqnarray*}
h(x)
&=&\E^x(e^{tH_A};H_A\geq1)+\P^x(H_A=0)\\
&=&\E^x(\E^xe^{t+tH_A\circ\Theta}|\F_1);X_0\notin A)+\P^x(X_0\in A)\\
&=&e^t\E^x(\E^xe^{tH_A\circ\Theta}|\F_1);X_0\notin A)+\P^x(X_0\in A)\\
&=&e^t\E^x(\E^{X_1}e^{tH_A};X_0\notin A)+\P^x(X_0\in A)\\
&=&e^t\E^x(h(X_1);X_0\notin A)+\P^x(X_0\in A)~.
\end{eqnarray*}
Hence if $x\notin A$: $Ph(x)=e^{-t}h(x)$. If $x\in A$ then both sides equal $1$.
2. In this case it follows that
$$
\forall |x| < p:\quad
h(x+1)+h(x-1)=2e^{-t}h(x)
\quad\mbox{and}\quad
h(\pm p)=1~.
$$
3. This homogeneous difference equation has the general solution
$$
h(x)=ae^{i\vp x}+be^{-i\vp x}
$$
for $a,b\in\C$. Now $h(\pm p)=1$ implies
$$
1=ae^{i\vp p}+be^{-i\vp p},
1=ae^{-i\vp p}+be^{i\vp p}~.
$$
Solving for $a$ and $b$ we get:
$$
a=b=\frac{\sin(\vp p)}{\sin(2\vp p)}=\frac{1}{2\cos(\vp p)}~.
$$
Hence
$$
h(x)=\frac{\cos(\vp x)}{\cos(\vp p)}~.
$$
Now $\vp < \pi/2p$ i.e. $e^{-t}=\cos\vp > \cos(\pi/2p)$, i.e.
$$
t < \log\frac1{\cos(\pi/2p)}~.
$$