Debouncer with LED inside
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. D1 is normally on. When SW1 is pressed D1 turns off and D2 turns on.

Note that the input will fall to 0.7 V minimum. Check that this is OK. I've changed the R values so it will affect your debounce timing. Re-calculate those too.



I've split R1 in two: R1 + R4. This raises the INPUT voltage by \$ \frac {V_{cc}-V_{D1}}{2} \$ to address Dave Tweed's comment.









simulate this circuit

Figure 2. A better design?

How it works:


With SW1 open 'B' is pulled high via R2 and D2. Meanwhile D1 is lit via R1. VA will be whatever Vf of the LED is so D3 will be reverse biased.
When S1 is pressed 'B' will be pulled to 0 V. D3 will now be in parallel with D1 but it's 0.7 V forward voltage will steal the current from D1 so it will turn off.