I think that trying to use RMS values of voltage and current for this is not going to work. Imagine shifting the current waveform 4ms later; neither the RMS voltage nor the RMS current would change at all, but the power drawn would change by an order of magnitude.
The instantaneous power drawn by your circuit is V * I. In any small time dt, the energy consumed will be V * I * dt. The energy consumed in 1s, the power drawn, will be the integral of V * I * dt from T=0 to T=1s. You could compute this directly from the sample values in your excel spreadsheet. At each time sample, multiply the instantaneous voltage by the instantaneous current, and that gives the instantaneous power drawn. Multiply that by the sample interval, and that is the energy consumed in that sample interval. Add all those up over an AC cycle, and multiply by the number of cycles per second and that is the energy drawn per second, otherwise known as the power.
Looking at the scope traces, the current drawn by the circuit is usually 0. Once per AC half-cycle, the current increases to about 90mA very quickly, then drops linearly to 0 over about 820us. It's a 60Hz circuit so it does this every 8.3ms. When the circuit is drawing current, the voltage is more-or-less constant at 170V. That's an average current of 45mA over the 820us at 170V = 7.65 W, but it only takes this power 1/10 of the total time, so the final power consumption is 0.76 W.
In my experience, the probability of wiring up a current probe backwards is exactly 0.5!
Ignore the capacitors, cable etc. and just consider the resistors. From probe tip to ground you have 9MΩ in series with 1MΩ - that's 10MΩ! You can verify this by measuring the resistance between probe tip and ground with a multimeter (on 20MΩ scale).
Of course this only applies to DC. At higher frequencies the impedances of the capacitors and cable become significant. At 50MHz the reactance of a 10pF capacitor is about 300Ω.
Best Answer
To see phase shift you need to compare two separate wave forms. Looking at one wave form won't tell you anything.
The two wave forms to compare are current and voltage. Seeing the voltage is easy, like you said, just connect the probe across the inductor. Seeing the current is more difficult because the oscilloscope can't "see" current. There are two options, use a current transducer to convert the current to voltage or use a resistor to do the same thing. Ohms law says that current through the resistor can be seen as a voltage drop across the resistor.
Be careful, don't connect the oscilloscope grounds in different places. That's a fast way to burn up your scope. The problem with that is the wave forms will have an additional 180 degrees phase shift. Just compensate for it when you do the calculations.
simulate this circuit – Schematic created using CircuitLab