= Solution
Lift $u$ through the <exponential function>[complex exponential] by setting
$$
v(w)=u(e^w),\qquad w\in\mathbb C.
$$
The <conformal invariance of harmonicity> makes $v$ harmonic. Since $e^{w+2\pi i}=e^w$ and $u(2z)=u(z)$,
$$
v(w+2\pi i)=v(w),
\qquad
v(w+\log2)=v(w).
$$
Thus $v$ is bounded on the compact rectangle
$$
\{s\log2+2\pi it:0\leq s,t\leq1\},
$$
and its two <period lattice>[periods] make it bounded on the whole plane. The <harmonic Liouville theorem> makes $v$ constant. Since the exponential map is surjective onto $\mathbb C^*$, $u$ is constant.
For the requested counterexample after deleting a countable set, take
$$
S=\{2^m:m\in\mathbb Z\}
$$
and let $\wp_\Lambda$ be the <Weierstrass elliptic function> for the lattice
$$
\Lambda=(\log2)\mathbb Z+2\pi i\mathbb Z.
$$
Define
$$
u(z)=\operatorname{Re}\wp_\Lambda(\log z),
\qquad z\in\mathbb C^*\setminus S.
$$
Changing a branch of the logarithm adds $2\pi i$, a period of $\wp_\Lambda$, so $u$ is well defined. Its poles project precisely to $S$, and the <conformal invariance of harmonicity> makes $u$ harmonic away from them. The other period gives
$$
u(2z)=\operatorname{Re}\wp_\Lambda(\log z+\log2)=u(z).
$$
Moreover, $z\notin S$ implies $2z\notin S$. Finally, $u$ is nonconstant because as positive real $z\to1$,
$$
\wp_\Lambda(\log z)\sim\frac1{(\log z)^2},
$$
which is real and unbounded. This is a <scale-periodic harmonic function from an elliptic function>.
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