 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 $3/(2z-1{)}^2\ne 0$.

For $z=x+iy$,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 $A=1$, $B=1/\log 2$. By conformal invariance of harmonicity,It has the required boundary values, and the maximum principle for harmonic functions proves uniqueness among solutions continuous on the closed region.