> diodes and the 100uF capacitor form a pair of half-wave envelope detectors?
Kinda.
> why does the network feedback to the signal output?
Not "feedback". (The upper opamp ideally ignores external stuff forced on its output.)
Think "push pull voltage doubler". Upper opamp makes a signal. Lower opamp makes an inverted signal. The difference "could" be *double* voltage. And we are short of voltage, with a bunch of diodes and just a 9V supply and lossy opamps.
However the lower opamp has gain of - 5. Since the LED voltage is higher than a happy "guitar level output", some gain is needed. However that means the push-pull action is not double, only 6/5 or 1.2
> Would running it to ground still allow the circuit to work
Yes, at 5/6 drive to the LED.
> measuring the resistance between the -ve and output terminals of the first opamp, and although I was getting funny negative resistance readings (?!), the resistance quite obviously changed when I struck a note...
Measuring in-circuit, especially around an opamp, is likely to read wrong.
Measuring "resistance" with large AC (audio) signals bopping around is likely to confuse your ohm-meter.
As a General Rule, when trying to read Ohms, your meter should be the ONLY power source in the circuit. No battery/wart, no signal.
In *this* case, there is a workaround. Disconnect the LDR side of the opto. Read that with ohm meter. Put a light-TIGHT cover over (or work in the dark). Should be way-high. (Over 100K, possibly so over 1meg that a DMM will read "---".) Run signal through. Heavy strums should drive the resistance down below 10K.