Really basic electeonics

Welcome.
Assuming you are listening monophonic, The selector you are using is a (six pin) double-pole-double-throw DPDT that each 'pole' routes amplifier 1 or amplifier 2 towards the headphones.
That is, centre pin to headphones, one end to amplifier 1 and the other end of the same switch section to amplifier 2 output.


IGNORE the triangles and their lines above - image borrowed from the web.
Circuit should be altered to :

NC1 and NO1 are your amplifiers outputs 1 and 2 positives. Com1 is your earphone positive.

NC2 to LED1 one; NO2 to your LED2. Com2 to +3V. All negative common grounds together for earphone, amplifiers outputs and LEDs.

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Can be done also by replacing your 6 pin selector a with a 9 pin one that has an extra set of switches 3PDT triple-pole-double-throw for the extra set to drive the leds, plus, the 3V supply for them to lit.

Note, your LEDs may need a series resistor.

Mike.

Thank you for the info. Actually, my connectors are stereo. Forgot to specify. Does this change anyth8ng?

Hi.
I do not know of stereophonic guitars. Each of your amplifiers should be monophonic. Using stereo connectors do not make a signal stereo. Perhaps binaural.
Headphones if stereo, can have both channels connected-merged at their jack for your single channel signal.
Then, from post #2 above, Com1 goes to both L+ and R+ merged headphone wires.

Thank you very much for the answers . I will try this . Thank you again