If DC is present it gets amplified along with the AC of interest. With any significant gain the DC will drive the output hard against either power rail. That prevents the AC giving any useful output.
With Relay1 contacts as shown, the 555 supply is shorted to ground, so the 555 won't then function. Although the relay coils should connect to Vcc the contacts shouldn't. They should instead connect to the 555 output pin.
You should add reverse-biased diodes across the relay coils (unless they...
If you get a replacement filament bulb, I'd suggest wiring a low value resistor in series with it to prolong its life without significantly reducing the brightness.
That depends a lot on your skills. You would have to dismantle the wheel (is it screwed or bonded together?), identify which USB wire/circuit-trace goes where, unsolder the damaged socket, then solder the wires/traces (extended if necessary) to a new socket.
One thing to consider with LEDs is beam angle. They are usually quite directional, whereas a filament bulb has a broad spread of light. So you may well need either a LED array or an effective light reflector/diffuser.
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The circuit is presumably copyrighted, patented or otherwise protected as Intellectual Property (IP), so you would need to get the IP rights holder's permission to go into production.
The circuit as is does not have the resistors defining specific voltages on the USB data lines...
1A is a helluva current for something intended to run from a 9V battery. What exactly does the battery power? A pickup preamp, an amplifier with a loudspeaker, or just a headphone amplifier?
Aarghhh, I missed that :oops:. Good catch.
So, three digital outputs suitably driven, plus some 1A rated output transistors (BJT or MOSFET), plus a couple of diodes will do the trick, and PWM isn't suitable.
If you check the online catalogue of any major electronics components distributor (such as Farnell, Digikey, Mouser, RadioSpares etc) you will see the various parameters/units used for selecting/characterising components.