= Solution
Use the <Fourier transform> convention $\widehat u(k,y)=\int_{\mathbb R}u(x,y)e^{-ikx}\,dx$ and inverse factor $1/(2\pi)$. <Laplace equation> gives $\widehat u_{yy}-k^2\widehat u=0$. Selecting its decaying Fourier branch gives $\widehat u(k,y)=\widehat u_0(k)e^{-|k|y}$, the standard bounded solution for bounded continuous boundary data. The inverse transform of the multiplier is the <Poisson kernel for the upper half-plane>:
$$
\frac1{2\pi}\int_{\mathbb R}e^{ikx-|k|y}\,dk
=\frac1\pi\int_0^\infty e^{-ky}\cos(kx)\,dk
=\frac{y}{\pi(x^2+y^2)}.
$$
The <convolution theorem> therefore gives the <Poisson integral>
$$
\boxed{u(x,y)=\int_{\mathbb R}u_0(t)\frac{y}{\pi[(x-t)^2+y^2]}\,dt.}
$$
For suitably regular boundary data, this kernel is an approximate identity as $y\downarrow0$, recovering $u_0$.
For $a>0$, contour integration gives $\widehat u_0(k)=(\pi/a)e^{-a|k|}$. Multiplication by $e^{-y|k|}$ and inversion yield
$$
\boxed{u(x,y)=\frac{y+a}{a[x^2+(y+a)^2]}.}
$$
The stated form assumes $a>0$; if $a<0$, replace it by $|a|$ because the boundary function only depends on $a^2$. The case $a=0$ has singular boundary data and is outside this regular calculation.
The transform argument requires a growth condition. A <zero-boundary harmonic function with pointwise vertical decay> shows that the printed conditions alone do not ensure uniqueness:
$$
v(x,y)=\operatorname{Im}e^{(x+iy)^2}=e^{x^2-y^2}\sin(2xy)
$$
vanishes at $y=0$ and tends to zero as $y\to\infty$ for every fixed $x$. It can be added without changing those conditions, but its growth in $x$ excludes it from the bounded solution class. The displayed Poisson integral is the standard bounded solution for regular bounded boundary data; pointwise vertical decay by itself cannot justify a uniqueness claim.
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