Another approach:
Claim: There exists $f=u+iv$ holomorphic in $U$ such that as $z\to 0$ in $U,$ $u(z) \to -\infty,$ $v(z)\to 0.$
Suppose the claim is proved and we have such an $f.$ Then the function $-if(-1/z) = v(-1/z) -iu(-1/z)$ is holomorphic in $U.$ As $z\to \infty$ within $U,$ $-1/z\to 0$ within $U.$ Hence $v(-1/z) \to 0$ and $-u(-1/z)\to \infty.$ Thus we have a counterexample.
Proof of claim: Consider the functions
$$f_n(z) =\log (z+i/e^n) = \log |z+i/e^n| + i\text {arg }(z+i/e^n),\,\,n=1,2,\dots,$$
where $\log $ denotes the principal value logarithm. These functions are holomorphic in $U$ and are uniformly bounded on compact subsets of $U.$ For $z\in U,$ define
$$f(z)=\sum_{n=1}^{\infty}\frac{f_n(z)}{n^2}.$$
By Weierstrass M, the series converges uniformly on compact subsets of $U,$ hence $f$ is holomorphic in $U.$ Writing $f=u+iv,$ we have
$$u(z) = \sum_{n=1}^{\infty} \frac{\log |z+i/e^n|}{n^2}.$$
Note all summands are negative on $\{z\in U: |z|<1/2\}.$ Thus for any $N,$
$$\limsup_{z\to 0} u(z) \le \lim_{z\to 0} \sum_{n=1}^{N} \frac{\log |z+i/e^n|}{n^2} =\sum_{n=1}^{N} \frac{\log |i/e^n|}{n^2} = - \sum_{n=1}^{N} \frac{1}{n}.$$
Since $N$ is arbitrary, we see $\lim_{z\to 0} u(z)=-\infty.$ (Just to be clear, these limits are taken as $z\to 0$ within $U.$)
As for $v(z),$ we have
$$v(z) = \sum_{n=1}^{\infty} \frac{\text {arg }(z+i/e^n)}{n^2}.$$
This series actually converges uniformly on all of $U.$ Thus
$$\lim_{z\to 0} v(z) = \sum_{n=1}^{\infty} \lim_{z\to 0}\frac{\text {arg }(z+i/e^n)}{n^2} = \sum_{n=1}^{\infty} \frac{\text {arg }(i/e^n)}{n^2} = \sum_{n=1}^{\infty} \frac{\pi/2}{n^2}.$$
We're not quite done: Letting $c$ denote the last sum, we see $f(z)-ic$ has the claimed properties.
Previous answer: I'm pretty sure this is false. Define $V=\{x+iy: x>0, |y|<x^2.$ Then $V$ is simply connected, so there is a conformal map $g:U\to V.$ And we should be able to arrange things so that $z\to \infty$ in $U$ iff $g(z)\to 0$ in $V.$ Now define
$$h(z) = -i\log g(z)= \text {arg }g(z) - i\ln |g(z)|.$$
As $z\to \infty$ in $U,$ $\text {arg }g(z) \to 0.$ That's because $g(z)\to 0$ tangent to the real axis. But the conjugate function $-\ln |g(z)| \to \infty.$
ΗΙΝΤ
$P(x),e^{-2\pi y||x||} \in L^p,\forall p \in [1,+\infty]$ since $y>0$.
You can see that by using the polar coordinates formula: $$
\int_{\Bbb{R}^n} f(x) dx = \int_0^\infty \left( \int_{S^{n-1}} f(re)d\sigma(e) \right) r^{n-1} dr.
$$
where $\sigma(E)$ is the surface borel measure on the unit sphere.
In this case your function is radial so the integration won't depend on $e \in S^{n-1}$ thus things are easy especialy since your function is non-negative.
Best Answer
When you say you don't see how to use CGT here it's not clear whether you don't see what needs to be done to apply CGT to the problem or you understand what needs to be done but don't see how to do it. Regarding what needs to be done:
Assume (i). We can then define a linear map $T:L^p(\Bbb R)\to L^p(\mu)$ by $$Tf(x+iy)=f*P_y(x).$$According to CGT, to show $T$ is bounded you can assume that $f_n\to f$ in $L^p$ and that $Tf_n\to u$ in $L^p(\mu)$, and you then need to show that $Tf=u$.
Hint regarding how to show $Tf=u$: If $1/p+1/p'=1$ then $P_y\in L^{p'}(\Bbb R)$. So $f_n\to f$ in $L^p$ implies...
Edit: Having revealed all in a comment I may as well give the argument here: We have $$|Tf_n(x,y)-Tf(x,y)|\le||f_n-f||_p||P_y||_{p'};$$hence $Tf_n\to Tf$ pointwise. Since $Tf_n\to u$ in $L^p(\mu)$ some subsequence tends to $u$ a.e.$[\mu]$; hence $Tf=u$ a.e.$[\mu]$.