I made a quick paint drawing of how it would be connected.
The IN4001 diodes drop about 0.7, but it will drop more as the power draw increases(and a bit less if the load is very low). I would guess only 1 diode would be required for this application, but if you start with 2, you can test the voltage and see what you get.
This is only 1 of many ways you can do this. As mentioned above you could use a zener diode regulator(2 components) or even a linear or switching regulator(regulators have the advantage of allowing a wide range of input voltages.), but for this project, I do not think it is required. If you knew the power consumption of your lights, you could have used a single resistor to "eat" the extra voltage.
You have 2 power supplies. 1 that is 5-6 volts @ 500ma. This power supply does not appear regulated and will most likely deliver closer to 6 volts under the light load of your leds. The LG power supply lists 5.1 and seems to be regulated. I think I would use the LG power supply.
To get power out of these units you will either have to cut the plug off or get a matching plug. Do not cut the plug off if you still need to use those power supplies.
In my illustration, you would connect to the existing battery compartment. This leaves the strobe mode and on and off switch functional. Connect negative to the far left top battery connector and positive to the lower right connector.
Once connected, before powering up(power supply plugged in and led's turned off), check the voltage with a meter set to DC(check at the newly connected battery terminals). Turn it on and check again. A small drop is to be expected as the leds will draw power once switched on. As long as you are under 4.5 volts, you are good to go. If the voltage is lower, you may notice dimmer light from the leds. This should not hurt the device or else dead batteries would hurt it.
If you have questions. just ask. I will try to make a demonstration picture a bit later today.
Here is a practical setup for testing. I have about 200ma of leds when run from 4.5 volts. The blue one is because I had no white with long leads left.
I placed the 2 IN4001 diodes in series with the + side. This causes about 1.54 volts to be dropped leaving about 3.46 volts. LEDS are dimmer, but light and will last forever.
With a single diode the voltage drop was about 0.8v leaving about 4.2 volts. leds are slightly dimmer, but with batteries the voltage had to drop overtime anyway.
I hope this gives you an idea of what I was talking about. You can also get USB or other plugs that can be mounted on the light. This would give a clean finished look to the project. I also would not use diodes to drop lots of voltage. if you had 12 volts, a switching regulator would be best(much less heat than a linear regulator).
EDIT first image has been re-added. I just fluke noticed it does not work on some mobile devices.