From diagram 3, I could think of two possibilities:
The microphone interface is polling for an existence of a mic periodically and turning the line off when not found. Probably not correct.
The mic input is doubled as another digital interface. Some phones have a way to detect whether the plugged in ear phones are manufacturer's original models or not, for example. If such a earphone pair is detected, the internal audio equalization may be tweaked to produce best possible fidelity.
The pulse shows narrow, I could not sit and figure out it's width, but it does look like a start bit for me. Assuming it is a start bit, it is probably followed by, say some bits and a stop bit - all are zeroes (like a UART). Beyond that, the controller probably is expecting a response.
The first diagram probably is the mic decoupling capacitor acting as weird RC filter with oscilloscope input impedance. Open drain outputs usually require a remote termination. That's another way to detect if remote is connected, somewhat like USB.
Which is the phone on the other end? I might not have access to the same model and I definitely don't have a similar cable handy to figure exactly out what is happening.
It might be interesting, though, to try sending an UART response on the line, taking the high pulse width as bit time: wait for ten or 11 bits and send high start bit and probably try some combinations. Even if we get a response, it might take a while to figure out the protocols etc. I would advise extreme caution, though - if it is a dangerous interface like a firmware backdoor, a trial sequence could potentially erase the flash.
You are very likely right in your suspicion of it being the cable.
Cheap cables from who knows where often use things like AWG34 wires, or in this case it might be coaxial with up to a whopping AWG30 center wire.
What you could try is adding a LOAD to it and measuring what comes out, for example a 2.5Ohm 10W resistor.
It can be made of 40 1Ohm 1/4Watt resistors by making 4 strands of 10 resistors in series and then connecting the ends together, or by connecting all strands at all points:
simulate this circuit – Schematic created using CircuitLab
The best solution is to find a cable that's ready made for 2A with low loss.
The next best one is to find a loose connector for each end and a wire with low resistance and connect them up all fresh. (The barrel connector will be easy, 100's of factories around the world make them at 20cents retail prices)
It's presumable that the device will still work with 4.7V, so the maximum loss in the cable itself will be limited by the plugs and the wire, I'll assume the connectors waste half the maximum voltage drop, because without specifics we can never be certain anyway. So that leaves 0.15V for the wire, which comes down to:
R = V/I = 0.15V/2A = 0.075Ohm
That's for the total wire, but you need a positive and a negative, so one wire can be up to 37mOhm. You can do two things: Estimate your length requirement, lookup the resistivity per meter of copper wires (there's a million tables on Google images relating AWG to Ohms per meter). Take a 25% margin and order the cable you need. But that does have the risk of the wires being too fat for the connector shell - some DIY required.
You can also order something in the range of AWG22~24, measure the voltage drop at 2A across 1m and then calculate the maximum length of that wire.
The next-next best thing is using connectors of existing cables and splicing a thicker wire in, it'll be easier, but the little ends of your cable will still be somewhat limiting, so you need to make them as short as possible.
"+5V" is the (or is connected to the) positive supply from the USB host, which should be 5V. "-5V" is mislabeled and should actually be the supply return, commonly called "ground". "D+" and "D-" are the D+ and D- USB data channels.