Limiting the current powering a circuit
If the LED bar works well when its terminals are have 4.61V across them, then you should shoot for that as a target.
To get your 2A 5.1V battery bank to supply the LED bar with 4.61V you need to drop 5.1-4.61= 0.49V.
If the current through the LED bar is 1.17A when the voltage is 4.61V, you need a 0.49V/1.17A = 0.419\$\Omega\$ resistance.
The power such a resistance will use is 0.49 x 1.17 = 573mW. I would go with a 1 watt resistor, or multiple resistors with lower power ratings in parallel, or series (adjusting the resistance values as appropriate). Yes, a resistor dissipating 573mW will get warm to the touch, but if properly rated, and exposed to air, it should not be in danger of failure, nor of starting a fire.
There are sub-Ohm resistors available.