So, the battery, if fully charged, could support the system for 10 days without beeing recharged.
The solar cell produces 5.5v, 120mA.
Even if the 1.8 volts extra voltage get entirely lost when reducing voltage in order to charge the 3.7V battery, and in spite of the continuous drain from the GPS, you would still get 115mA to recharge the battery, with the solar cell at its peak power.
With only one hour of full sunlight each day, you run your GPS, and also send the extra 115mAh to recharge the battery. These 115mA are exactly what you need to supply 5mA to the GPS in the other 23 hours of day (23h x 5mA = 115mA).
The correct calculation should be:
0.005A x 3.7v = 0.0185 W power consumption of GPS device
0.12 A x 5.5v = 0.66 W theoretical maximum power of solar cell
0.66 - 0.0185 = 0.6415 W hourly power available to recharge the battery
3.7v*1.2Ah/0.6415 = 6.92 hours of high power direct sunlight to fully charge
Even if you start with a empty battery, the solar cell outperforms the device in a ratio of 35:1, so, there's a lot of energy to keep charging the cell, and the cell gets energy enough to keep the GPS working until the next sunlight time, when the charging starts again. This configuration seems very stable, IMHO.
So, even with many unavoidable power losses in conversion and heating and etc. and even with sub-optimal sunlight (but remember that the battery can cope with 10 days without sun), it seems to me that the system that you described would run until the end of battery's lifetime.