Input impedance oscilloscope
Regardless of what the scope's impedance is, it still shows you the voltage at its input.  The power the scope is taking from that signal is V2/R.  In your case that comes out to 97 mW, as you calculated.

So this is a long way of answering "yes" to your question.

Note that the power going into the scope is not the same as the power put out by the voltage source inside the signal generator (V1 in your diagram).  It's actually double the 97 mW since R1 is also dissipating 97 mW. However, that's all stuff that happens inside the black box, and is irrelevant from a outside point of view.