Here is a vectorized approach (as in "look ma, no coordinates!"). Make sure you have the chain and product formulas written down in convenient form (they help with other calculations too):
$$\nabla(\varphi(u)) = \varphi'(u) \nabla u\tag{1}$$
$$\Delta(u) = \operatorname{div} \nabla u \tag{2}$$
$$\operatorname{div} u \mathbf F = \nabla u\cdot \mathbf F
+ u \operatorname{div} \mathbf F\tag{3}$$
$$\Delta(uv) = u\Delta v+v\Delta u+2\nabla u\cdot \nabla v \tag{4}$$
Your function is $uv$ with $u=(1-\|x\|^2)$ and $v=\|x-\xi\|^{-n}$.
We have
$$\nabla u = -2 x,\quad \Delta u = -2n $$
Using (1):
$$
\begin{split}
\nabla v &= -n \|x-\xi\|^{-n-1}\nabla \|x-\xi\| \\
&= -n \|x-\xi\|^{-n-1}\frac{x-\xi}{\|x-\xi\|} \\
&= -n \|x-\xi\|^{-n-2}(x-\xi)
\end{split}
$$
Using (2) and then (3):
$$
\begin{split}
\Delta v &= -n \operatorname{div} ( \|x-\xi\|^{-n-2}(x-\xi)) \\
&= -n (-n-2) \|x-\xi\|^{-n-3}\frac{x-\xi}{\|x-\xi\|}\cdot (x-\xi)
-n \|x-\xi\|^{-n-2} n \\
& =2n\|x-\xi\|^{-n-2}
\end{split}
$$
Finally, combine the results using (4). For convenience, I multiply the Laplacian by $\|x-\xi\|^{ n+2}$:
$$
\begin{split}
\|x-\xi\|^{ n+2}\Delta(uv)
&= -2n \|x-\xi\|^{2} + (1-\|x\|^2) 2n +4nx \cdot (x-\xi) \\
& = 0
\end{split}
$$
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 $-\ln |g(z)| \to \infty.$
Best Answer
Say $u_y(t)=u(t+iy)$. If you look at the Fourier transform of $P(.,y)$ you see that $$\widehat u_y(\xi)=e^{-y|\xi|}\widehat f(\xi).$$
Hence $$||u_y||_2^2=\int e^{-2y|\xi|}|\widehat f(\xi)|^2\,d\xi,$$so $$\int_0^\infty||u_y||_2^2\,dy=\int\frac1{2|\xi|}|\widehat f(\xi)|^2\,d\xi;$$ so $u\in L^2$ if and only if that last integral is finite.
(This even makes sense. The reason $||u_y||_2$ should tend to zero is cancellation in the integral; saying $\widehat f(0)=0$ says $\int f=0$, cancellation.)