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
METHODOLOGY $1$: COMPLEX ANALYSIS
Let $Q_t(x)=\frac{x}{\pi^2(x^2+t^2)}$. Note that the Fourier Transform of $Q_t(x)$ is given by
$$\begin{align} \hat Q_t(\xi)&=\int_{-\infty}^\infty \frac{x}{\pi^2(x^2+t^2)}e^{i\xi x}\,dx\\\\ &=\begin{cases} 2\pi i \text{Res}\left(\frac{z}{\pi^2(z^2+t^2)}, z=it\right), &\xi>0\\\\ -2\pi i \text{Res}\left(\frac{z}{\pi^2(z^2+t^2)}, z=-it\right), &\xi<0 \end{cases}\\\\ &=\frac{i}{\pi}\text{sgn}(\xi)e^{-|\xi t| } \end{align}$$
where we used contour integration and the Residue Theorem to arrive at the coveted result.
METHODOLOGY $2$: REAL ANALYSIS
Alternatively, we can evaluate the Fourier Transform using real analysis. Proceeding, we see that
$$\begin{align} \hat Q_t(\xi)&=\int_{-\infty}^\infty \frac{x}{\pi^2(x^2+t^2)}e^{i\xi x}\,dx\\\\ &=\int_{-\infty}^\infty \frac{(x^2+t^2)-t^2}{\pi^2x(x^2+t^2)}e^{i\xi x}\,dx\\\\ &=\frac i \pi \text{sgn}(\xi)-\frac{t^2}{\pi^2}\int_{-\infty}^\infty \frac{1}{x(x^2+t^2)}e^{i\xi x}\,dx \end{align}$$
Differenting twice, we find that
$$\begin{align} \hat Q_t''(\xi)&=\frac{t^2}{\pi^2}\int_{-\infty}^\infty \frac{x}{x^2+t^2}e^{i\xi x}\,dx\\\\ &=t^2\hat Q_t(\xi) \end{align}$$
Solving the ODE, $\hat Q_t(\xi)=C_1e^{|t|\xi}+C_2e^{-|t|\xi}$.
Now, as $t\to \infty$, $\hat Q_t(\xi)\to 0$. Hence, we must have $\hat Q_t(\xi)=C^+e^{-|t|\xi}$ for $\xi>0$ and $\hat Q_t(\xi)=C^- e^{|t|\xi}$ for $\xi<0$.
We also see that $\hat Q_t'(0)=-i|t|/\pi$ and $\hat Q_t'(0)=- \text{sgn}(\xi) |t|C^\pm$. Therefore, $C^\pm=\frac{i}\pi \text{sgn}(\xi)$.
Putting it all together, we have
$$\hat Q_t(\xi)=\frac{i}{\pi}\text{sgn}(\xi)e^{-|t\xi|}$$
as expected!