Hi @mfikes - That is a great, simple test circuit. Thank you for sharing the great screen captures and test circuit schematic. That 555 test circuit likely has rise times measured in 10s of nanoseconds, which is pretty unfriendly, too 
I suspect that this fast transient is causing the JS110 analog front-end opamp to misbehave. The JS110 uses the MAX4239, which is a chopper opamp. It definitely does not like being driven into overload. From page 4 of the datasheet:
In the screen captures, your Joulescope was in current_range 4, which uses a 11.1 Ω shunt resistor. From section 8.5 on page 12 of the Joulescope User’s Guide:
With your circuit, 3.3 V / 22 Ω = 0.15 A. Your Joulescope sees 11.1 Ω * 0.15 = 1.6V. The opamp has a gain of 11 and saturates around 0.31 V, which is only 28 mA, so this circuit is definitely driving the MAX4239 into overload. While recovering from overload, the MAX4239 typically oscillates with alternating pulses between 10 kHz and 15 kHz, which is the chopper amp frequency.
Do you want to test my hypothesis that the MAX4239 is misbehaving? If so, you can modify your circuit to extend the pulse duration. I would expect that you will see alternating + & - ringing at the chopper amp frequency, with pulses somewhere between 66 and 100 µs. You can add dual markers to measure the duration.
We normally recommend that the target contain at least 10 µF of capacitance to reduce the system bandwidth and slow down those fast edges. In your test circuit, you would place this capacitance in parallel with the 22 Ω resistor R1 and Q1. This forms a RC filter. The R is actually the supply resistance and the JS110 shunt resistance. This slows the rate at which VCC drops when Q1 switches on. You can compute the time to saturation using I(t) = Imax * (1 - e ^ (t / τ)) [Wikipedia]. So t = τ ln (1 - I(t) / Imax) = 11.1 * 10e-6 * ln(1 - 0.028/0.15) = 23 µs, which is plenty of time for the JS110 to react and switch to current range 0.
Does this make sense?