Since the general solution of Laplace’s equation is $u(x,y)=F(x+iy)+G(x-iy)$ , for matching the only condition $u(x,0)=\sqrt{x}$ , you can obviously take $F(x)=\dfrac{\sqrt{x}}{2}+f(x)$ and $G(x)=\dfrac{\sqrt{x}}{2}-f(x)$ so that you can get $u(x,y)=\dfrac{\sqrt{x+iy}+\sqrt{x-iy}}{2}+f(x+iy)-f(x-iy)$
Note that this solution not only satisfy on $-\infty<x<\infty$ and $0<y<\infty$ but also satisfy on $x,y\in\mathbb{C}$ .
We know that the exponential function gives a biholomorphism between an open strip parallel to the real axis with a width $\leqslant 2\pi$ and an angular sector. One angular sector that is particularly easy to handle, especially considering we want to reach the unit disk, is the upper half plane. We need a width of $\pi$ for that, so
$$f_1(z) = e^{\pi z}$$
gives a conformal mapping of $U$ to the upper half plane, and it maps the $y = 0$ part of the boundary to the positive half-axis, and the $y = 1$ part to the negative half-axis.
That looks promising, now we need a biholomorphic map between the upper half-plane and the unit disk that maps the positive half-axis to the part of the unit circle in the lower half-plane, and the negative half-axis to the part in the upper half-plane. That means it must map $\{0,\infty\}$, the remaining part of the boundary of the upper half-plane, to $\{-1,1\}$, the remaining part of the boundary of the unit disk. With the boundaries properly oriented, both regions lie to the left of their respective boundaries, so when the positive half-axis is traversed from $0$ to $\infty$, the lower unit semicircle is traversed from $-1$ to $1$, hence the biholomorphism must map $0 \mapsto -1$ and $\infty \mapsto 1$. The Möbius transformation
$$f_2(z) = \frac{z-i}{z+i}$$
is easily seen to be the solution to that problem.
We need to compose with $f_1^{-1} \circ f_2^{-1}$, so let's compute the inverses and compose:
$$f(z) = \frac{1}{\pi} \log \left(i\frac{1+z}{1-z}\right)$$
biholomorphically maps the unit disk to the strip $0 < \operatorname{Im} z < 1$, mapping the unit semicircle in the lower half plane to the real axis, and the upper unit semicircle to the line $\operatorname{Im} z = 1$.
Best Answer
Starting with the separation of variables solution, there is a way to sum the series $$ u(x,y)=\sum_{n=0}^{\infty}A_n\cos(n\pi x)e^{-n\pi y}. $$ $u(x,0)=f(x)$ determines the coefficients $A_n$: $$ f(x)=\sum_{n=0}^{\infty}A_n\cos(n\pi x) $$ $$ \int_0^1 f(x)\cos(n\pi x)dx = A_n\int_0^1\cos(n\pi x)^2dx $$ $$ A_n = \frac{\int_0^1 f(x)\cos(n\pi x)dx}{\int_0^1\cos^2(n\pi x)dx}. $$ Therefore, $$ u(x,y)=\sum_{n=0}^{\infty}\frac{\int_0^1f(x')\cos(n\pi x')dx'}{\int_0^1\cos^2(n\pi x')dx'}\cos(n\pi x)e^{-n\pi y} \\ = \sum_{n=0}^{\infty}2\int_0^1f(x')\cos(n\pi x')\cos(n\pi x)e^{-n\pi y}dx' \\ = 2\int_0^1f(x')\sum_{n=0}^{\infty}\cos(n\pi x')\cos(n\pi x)e^{-n\pi y}dx' \\ = 2\int_0^1f(x')\sum_{n=0}^{\infty}(\cos(n\pi(x'-x))+\cos(n\pi(x+x'))e^{-n\pi y}dx' \\ = 2\Re\int_0^1f(x')\sum_{n=0}^{\infty}(e^{in\pi(x'-x)}+e^{in\pi(x'+x)})e^{-n\pi y}dx' $$ Now it comes down to summing a geometric series, which I'll leave to you.