Without more numerical information it is only possible to suggest what is happening.
The resistance of a working $60\, \rm W,\, 230 \, V$ light bulb is approximately $880 \, \rm \Omega$ whilst when cold it is about $70 \, \Omega$.
So the resistance of a light bulb does change a lot with the voltage across (current though) it.
The voltage decreased slightly as expected.
Probably indicates the the power supply, voltage $V$, has an internal resistance $X$.
Without reasonable accurate figures the next bit is speculation but it does show that the current through the bulbs can increase by an "unexpected" amount.
Given that the current through one bulb was $0.04\, \rm A$ which is a long way from the rated current of $0.1\, \rm A$ I will assume/guess that the voltage across it was much less than its rating of $6 \, \rm V$.
For one bulb the current through it was $0.04 \, \rm A$ and assume that the voltage across it was $3\, \rm V$ (a guess) and when two bulbs were in parallel the current through one bulb was $0.09 \, \rm A$ and assume that the voltage across them was $1\, \rm V$ (another guess to make the number work).
Using these values one finds that the resistance of one light bulb when the current was $0.04\, \rm A$ was $75\,\rm \Omega$ and the resistance of one light bulb in parallel with another when the current was $0.09\, \rm A$ was $11\,\rm \Omega$.
The voltage of the supply was $3.6 \, \rm V$ and its internal resistance was $14 \,\Omega$.
Importantly, the power dissipated in one bulb when alone was $0.12 \,\rm W$ and when in parallel with another was $0.09 \,\rm W$.
This lowering of the power dissipated in the bulb with another one in parallel with it would result in a lower filament temperature and hence a lower resistance.
So perhaps there is scope for further analysis of the data already obtained but also further experimentation to include more than two bulbs in parallel?
However after reading the comments it sounds as though some more sensitive meters should be used?