Define a function $g = h'_x - ih'_y$. Since $h$ is harmonic (and hence $C^2$), $g$ satisfies Cauchy-Riemann's equations on $\Omega$. Indeed:
$$
(h'_x)'_x = h''_{xx} = -h''_{yy} = (-h'_y)'_y
$$
and
$$
(h'_x)'_y = h''_{xy} = h''_{yx} = -(-h'_y)'_x.
$$
In other words, $g$ is holomorphic on $\Omega$. Assume for a moment that $g$ admits an anti-derivative $G$ on $\Omega$. If $G = U+iV$, Cauchy-Riemann shows that $G' = U'_x-iU'_y$, but $G' = g = h'_x-ih'_y$, so $\nabla U = \nabla h$, i.e. $U=h+C$, and we may as well take $C=0$ by adjusting our choice of $G$. Hence $h = U = \operatorname{Re} G$.
Now, the problem is that $\Omega$ is not simply connected, so we can't guarantee that $g$ admits an anti-derivative. On the other hand, we know that $g$ has an anti-derivative on $\Omega$ if and only if $\int_\gamma g(z)\,dz = 0$ for every simple closed curve in $\Omega$, and it's not difficult to see that this condition is equivalent to the fact that the residues of $g$ at $0$ and $1$ should both be $0$.
Let $a_0 = \operatorname{Res}\limits_{z=0} g$ and $a_1 = \operatorname{Res}\limits_{z=1} g$, and put $u(z) = h(z)-a_0\log|z|-a_1\log|z-1|$.
Repeat the above argument to get a holomorphic function $f=u'_x-iu'_y$ on $\Omega$. It's straight-forward to verify that
$$
f(z) = g(z) - \frac{a_0}{z} - \frac{a_1}{z-1}
$$
so by construction, the residues of $f$ at $0$ and $1$ vanish, and from the discussion above, we are done.
This takes care of existence. For uniqueness, assume that $u(z) = h(z)-a_0\log|z|-a_1\log|z-1|$ and $\tilde u(z) = h(z)-\tilde a_0\log|z|-\tilde a_1\log|z-1|$ are two such functions. Then $u(z) = \tilde u(z) = (a_0-\tilde a_0) \log|z|-(a_1-\tilde a_1)\log|z-1|$ is the real part of a holomorphic function on $\Omega$. It's a standard exercise to check that this can only happen if $a_0 = \tilde a_0$ and $a_1 = \tilde a_1$.
