This question probably applies to other heavier metals as well. The only rationalization I can figure out is that lead in the +2 oxidation state (which is most common) is a borderline soft acid and prefers to take electrons from a soft sulfur instead of trying to react with a hard oxygen.
Is this the right direction? Thank you in advance.
Lead(II) and sulfide ions are large and polarizable, making them soft in the context of hard/soft acid/base theory (HSAB), which states that ions with similar hardness will form stronger bonds. The oxide ion is hard and therefore has a tendency to form weaker interactions with Pb(II).
HSAB theory has been discussed on Chem.SE in this question and this answer.
I guess that a long time ago there was a lot of $\ce{H2S}$ in the atmosphere and then the $\ce{H2S}$ could have reacted with that lead oxide to form $\ce{PbS}$?
On wikipedia it says (under Lead(II)oxide,reactions) :
The red and yellow forms of this material are related by a small change in enthalpy:
PbO(red) → PbO(yellow) ΔH = 1.6 kJ/mol
PbO is amphoteric, which means that it reacts with both acids and with bases. With acids,it forms salts of $\ce{Pb2+}$ via the intermediacy of oxoclusters such as $\ce{[Pb6O(OH)6]4+}$. With strong base, $\ce{PbO}$ dissolves to form plumbite(II) salts:[6]
$\ce{PbO + H2O + OH- -> [Pb(OH)3]-}$
It could have reacted with $\ce{H2S}$ following this reaction : (in water)
$\ce{Pb2+ + H2S -> PbS + 2 H+}$
