The way you possibly do it could do it is by getting a bit of breadboard and mount 3 x HE3621A0510 reed relays on it. HE3621A0510 reed relay has a min. coil operating voltage of 3.65v and switching current of 0.5A with a carry current of 1.2A so your 4.5v so should be enough to drive the coil in the relay.
Reed relays by design are silent in operation so no clicking sound in operation. At 12v 2w each bulb will pull approx. 170 milliamps
Connect GND of your original LED light circuit to pin 3 and 4.5v positive output to pin 2 to drive the coil of the relay then have a separate 12v power supply with the GND connected to the negative on the G4 bulb and then the positive of the 12v power supply connected to pin 4 and the positive of the G4 bulb connected to pin 1 of the relay.
With the original 4.5v circuit powering the LEDS it will now be powering the reed relays instead and be able to power the higher voltage bulbs from an separate 12v power supply.