Saving the state of the LED on illuminated momentary button

Don't listen to caveman.

^{simulate this circuit – Schematic created using CircuitLab}

Flip-flops can be synthesized from gates, although you may well decide to take Spehro's advice.

In the schematic, Switch 5 is a reset, and pressing it causes all 4 outputs (at the LED) to go high, turning off the LEDs. Pushing any of Switches 1 - 4 will cause the corresponding LED to turn on, and disable any other switch activations by way of the AND gates.

If you want to use discrete logic, such as 7400/74LS/74HC, etc, the total package count is 5 ICs - 1 7400, 2 7410 and 2 7411.

ETA - You'll notice that the LEDs are driven by being pulled down, with current coming from VCC. This is important if you decide to use 7400 or 74LS - those technologies do not source current well, but they sink current just fine. If you decide to go with a CMOS family such as 74HC, this will still work. Or, if you like, you can use the other output from each latch, the 2-input NAND gate, and drive the LED to ground (being careful to get the LED polarity right). Just a thought.

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When looking at the connections for common Cathode / Anode LEDs you can see that once you connect the common node appropriately (i.e. Vdd/Gnd) then you just need to drive the RGB lines High for Common Cathode and Low for Common Anode to illuminate the LEDs.

UPDATE: If (as OP states) you're prepared to wire-link the common node to Vdd or Gnd then you just need to make the RGB drive signals selectable positive/negative logic (i.e. 1= ON or 0=ON).

You can do this easily with an XOR gate or a MUX.

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Here the MUX selects line selects A or !A to so you can drive common Cathode / Anode respectively.

Looking at the XOR version you can see B is inverted if A=1 .

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