= Solution
On the specified <ellipse>, the semiaxes are $\sqrt2$ and $1$, so its elliptic coordinate is $u_0=\operatorname{arcosh}\sqrt2=\log(\sqrt2+1)$. Thus in the exterior,
$$
\boxed{\phi=\operatorname{Re}\log\left(z+\sqrt{z^2-1}\right)-\log(\sqrt2+1)}
$$
is a <harmonic function> with a homogeneous <Dirichlet boundary condition> on that ellipse. Choose the square root with $\sqrt{z^2-1}\sim z$ at infinity. Although a global complex logarithm requires a cut, its real part $\log|z+\sqrt{z^2-1}|$ is well defined throughout the exterior and gives the same $\phi$.
Expanding at large $|z|$ yields $z+\sqrt{z^2-1}=2z[1-1/(4z^2)+O(|z|^{-4})]$. Therefore
$$
\boxed{\phi=\log\frac{2\sqrt{x^2+y^2}}{\sqrt2+1}+O\bigl((x^2+y^2)^{-1}\bigr).}
$$
In particular the potential grows logarithmically in the distance, in every direction at infinity. It is the exterior solution; no <harmonic> continuation through the focal segment in the ellipse's interior is claimed.
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