An update. Using the fact that, in the Southern hemisphere, Mathematica azimuth zero is at local meridian North. As is common. With positive direction towards East. And so coincides with longitude zero and direction, when seen from the South pole..

moonGeoPositionNr2[datetime_DateObject] := Module[
    {latitude, longitude},
    latitude = First@Last@MoonPosition[datetime, CelestialSystem -> "Equatorial"];
    longitude = First@First@MoonPosition[GeoPosition@{-90, 0}, datetime];
    GeoPosition@{latitude, longitude}
  ]

theTime = DateObject[{2016, 2, 27, 18, 0, 0}, TimeZone -> 0]

moonGeoPositionNr2@theTime

GeoPosition[{-9.209, 145.40}]

Got carried away, and added date/time specification:

moonGeoPosition[datetime_DateObject] := Module[
    {longitude, declination},
    longitude = 180 - First@First@MoonPosition[GeoPosition@{90, 0}, datetime];
    declination =First@Last@MoonPosition[datetime, CelestialSystem -> "Equatorial"];
    {longitude, declination}
  ]

moonGeoPosition@DateObject[{2016, 2, 26, 22, 0, 0}, TimeZone -> 0]

{76.69, -6.414}

Awesome! There is a piece of code does the similar thing in the document page of MoonPosition --> Neat Example:

The CelestialSystem basically wraps the change of basis thing into one option.

Sorry, this still results in a visible reference from a point on earth. Getting to ground position is either a very complex proof or an already solved problem. Does anybody now how to get to the ground position. I did not ask this without trying for hours.