= Solution
The inner circle has centre $2/5$ and radius $2/5$. The pole $z=1/2$ of the <Möbius transformation> lies inside that removed disc, so the transformation is holomorphic on $D$, with derivative $3/(2z-1)^2\ne0$.
For $z=x+iy$,
$$
|w|^2=\frac{|z|^2-4x+4}{4|z|^2-4x+1}.
$$
On the outer boundary this is one. On the inner boundary, $|z|^2=4x/5$, and the ratio is four. More directly, inside the outer circle the numerator exceeds the denominator by $3(1-|z|^2)>0$, whereas outside the inner circle four times the denominator exceeds the numerator by $3(5|z|^2-4x)>0$. Thus the image is $1<|w|<2$. The inverse transformation is $z=(w-2)/(2w-1)$, so this is a bijective <conformal map> onto that annulus.
A radial <harmonic function> on an annulus has form $A+B\log|w|$. The two boundary values give $A=1$, $B=1/\log2$. By <conformal invariance of harmonicity>,
$$
\boxed{\phi(x,y)=1+\frac{\log\left|\dfrac{z-2}{2z-1}\right|}{\log2},\qquad z=x+iy.}
$$
It has the required boundary values, and the <maximum principle for harmonic functions> proves uniqueness among solutions continuous on the closed region.
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