Let $h(x,y)$ denote the probability that a Brownian motion started at $(x,y)$ hits $K = \{x < 1, |y| < 1\}$ before it leaves $D = \{|y| < 1\}$. That is, $h = 1$ on $K$, $h = 0$ on $\partial D$ and $h$ is harmonic in $D \setminus K$ (plus the usual continuity condition on regular boundary points of $D \setminus K$). The killed Brownian motion in $D$ conditioned to hit $K$ is the Doob $h$-transform of the usual killed Brownian motion in $D \setminus K$.

Now let $L = \{y = \tfrac{1}{2}\}$. The probability $f(x, y)$ that the conditioned process started at $(x, y)$ hits $K$ no later than it hits $L$ is an $h$-harmonic function in $D \setminus (K \cup L)$ which takes values $1$ in $K$ and $0$ in $L \setminus K$ and in $\partial D$ (plus the usual continuity condition). Therefore, $g(x, y) = f(x, y) h(x, y)$ is harmonic in $D \setminus (K \cup L)$, takes values $1/h(x, y) = 1$ on $K$ and $0$ on $K \setminus L$ and on $\partial D$.

Your question is: does $f(x, y) = g(x, y) / h(x, y)$ converge to zero as $|(x, y)| \to \infty$? Equivalently: does $g(x, y)$ converge to zero *faster* than $h(x, y)$ does?

It can be proved (using, for example, a boundary Harnack inequality argument) that $h(x, y) \approx e^{-\pi x/2} \sin(\tfrac{1}{2} \pi (y + 1))$ as $x \to \infty$, $|y| < 1$. Similarly, $g(x, y) \approx e^{-2 \pi x} \sin(2 \pi (y + 1))$ as $x \to \infty$, $y \in (-1, -\tfrac{1}{2})$, and $g(x, y) \approx e^{-2 \pi x/3} \sin(\tfrac{2}{3} \pi (y + \tfrac{1}{2}))$ as $x \to \infty$, $y \in (-\tfrac{1}{2}, 1)$. Thus, the answer is **yes**.

Of course, one can ask the same question for more general diffusions, as well as for more general $K$ and $L$. Whather there is a similar answer depends on what one knows about these objects: one needs some control over the behaviour at infinity of positive harmonic functions in $D \setminus K$ and in $D \setminus (K \cup L)$ at infinity.