Can I add an LED light to an existing device?
Assuming you can tap the the V+ and V- and get two free wires out of it, there is a very easy way to do it with SMD LEDs. But it requires some soldering skill.
If you have never soldered tiny SMD components, you will have to train a few times before but it's easier than it looks like.
-Find 20mA SMD LEDs, GREEN (eventualy blue or red, but green shine better. NOT yellow or orange). Any brand or no-brand is ok.
- one small prototype board to solder the LEDs on. 
- Find 10K ohms resistors. They can be through hole or SMD, at your convenience.
- Connect the anode of the led to V+, the cathode of the led to a 10K resistor, and the other side of the reistor to V-. (you can place the resistor before the led instead of after, makes no difference)

This works with SMD leds. with 10K resistor you are safe up to 36V. If the voltage is between 5 and 36V it's ok. you don;t even have to care what voltage it is. Make only sure it's DC.
If the LED is too dim, try a lower value. But not less than 4.7K for 12V and above. 

You can try with through hole (THT) "melon hat" LED if you are afraid of SMD job. But in this case you will have to choose the resistor more accurately else they won't shine.
They should work with a resistor between 450 ohms and 2.2 Kohms with 12V. 
The exact resistance can be calculated according to the voltage of the LED (usually 1.8 or 3V depending on the color) and the voltage of the power supply.
The formula is R = (Vsupply - Vled)/Iled Where Iled is usually 20 mA for indicator leds. This gives you the minimum value for the resistor for the best output. You can use higher values than given by the formula as long as the LED shines enough to be seen.
When using the optimal value given by the formula, and the voltage is 12V or more, you must chose a 1/4 watt resistor or more. When the resistance is much higher than the optimal value, like 10K as suggested, the power is so small dissipation that you don't need to worry about it.