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.
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 this question, which explains how to construct $f$.)
$v(z)=\Im f(z)$ 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 $\theta\in(0,\pi)$.
Take without loss of generality $a=0,\; b=\pi$, then $e^{-i(z+\pi)}$ maps your strip $0<z<\pi$ onto 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 $\infty$. 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 portion of the boundary and bounded in your strip.