If you want the positive answer you should state your last condition more carefully.
For example, add that $u$ is bounded, or that $u(x+iy)$ tends to $0$ as $y\to\infty$
UNIFORMLY with respect to $x$.

As you presently stated, the answer is negative. I sketch the construction of a counterexample.

1. There exists a non-zero entire function, real on the real line and such that $f(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta$. (See, for example, my answer to <a href="https://mathoverflow.net/questions/190837/entire-function-bounded-at-every-line/190875#190875">this question</a>, which explains how to construct $f$.)

2. $v(z)=\Im f(z)$ is a non-zero harmonic function in the upper half-plane, equal to $0$ on
the real line and $u(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta\in(0,\pi)$.

3. Take without loss of generality $a=0,\; b=\pi$, then $e^{-i(z-\pi)}$
maps your strip $0<z<\pi$ into the upper half-plane, with a removed half-disk.
So the function $w(z)=v(e^{-i(z-\pi)})$ is harmonic, zero on infinite sides
of your half-strip and satisfies the property at $\infty$:
$w(x+iy)\to 0$ as $y\to+\infty$ for every $x\in(0,\pi)$. Notice that this function is unbounded. To satisfy the last requirement, that $u(x)=0$ for $0<x<\pi$,
set $u(z)=w(z)-w_1(z)$, where $w_1(z)$ is the solution of Dirichlet problem matching
the boundary values of $w$ on the finite part of the boundary and bounded in your strip.