I would say that the short answer is yes, it is a decent measurement setup. The longer, and slightly more annoying answer is, as it often is in engineering, it depends on what accuracy you are after
Regardless however of the accuracy that we are after we need a model in order to say something about what we are measuring, and how well.
Let's first create a formal mathematical description of the quantity that we are trying to measure, if I have understood the question correctly, we are trying to find the transfer function between the power dissipation and the base emitter voltage or;
$$G(s)=\frac{V_{BE}}{P}$$
The following is an approximation to the Ebers–Moll model for a BJT, it is sufficiently accurate for now;
$$I_E=I_{ES}(e^{\frac{V_{BE}}{V_T}}-1)$$
We don't care about the emitter current but rather the base emitter voltage, and its dependence on V_T (the thermal voltage), so let's rearrange;
$$V_{BE}=V_T\ln{(\frac{I_E}{I_{ES}}+1)}$$
Since I_ES is assumed to be constant, and since I_E is kept constant, we see that indeed V_BE is proportional to V_T.
For V_T we have the equation;
$$V_T=\frac{kT}{q}$$
Where k is boltzmann's constant, q is the elemental charge (both are constants), and T is the temperature in Kelvin.
For now let's assume that the transfer function between the power dissipation and the junction temperature can be modelled as a single pole low pass filter as follows;
$$\frac{T}{P}=\frac{R_{th}}{1+\frac{s}{\omega_c}}$$
Where R_th is the thermal resistance and omega_c is the corner frequency
Combining all of the above we have the transfer function;
$$G(s)=\frac{V_{BE}}{P}=\frac{kR_{th}}{q(1+\frac{s}{\omega_c})}\ln{(\frac{I_E}{I_{ES}}+1)}$$
So if we assume that the thermal resistance is the same between your test setup and the final application then the right hand side of the equation above is just a constant and your measurements hold.
Now to tear it appart
Let's re-evaluate the approximations that we made before;
- We assumed that the simplified Ebers-Moll model was sufficiently accurate, is it?
Well the full Ebers-Moll model for the emitter current in a BJT is as follows;
$$I_E=I_S((e^{\frac{V_{BE}}{V_T}}-e^{\frac{V_{BC}}{V_T}})+\frac{1}{\beta_F}(e^{\frac{V_{BE}}{V_T}}-1))$$
Let's again solve for V_BE to get;
$$V_{BE}=V_T\ln{(\frac{\frac{I_E}{I_S}+e^{\frac{V_{BC}}{V_T}}+\frac{1}{\beta_F}}{\frac{1}{\beta_F}+1})}$$
We now see that the base emitter voltage is not just proportional to the thermal voltage V_T, but is also dependent on the base collector voltage V_BC, which in the case of your test setup is not a constant.
- We also assumed that the emitter current I_E was a constant, is it?
Well it's probably close enough, but in reality the voltage across the bottom resistor R_2, and hence the emitter current, is going to depend on the base emitter voltage itself, as well as R_3 and the base current. I am not going to calculate by how much right now as it's a harry bit of math work and because I think the error it introduces is fairly negligible, but it is worth noting that this causes the factor on V_T to be non constant.
- We assumed that the thermal resistance R_th was the same between your test setup and the final application, is this the case?
I don't know your test setup well enough to draw any conclusions in this regard, so I will leave that as a question for you.
- We didn't considder any parasitics such as collector-emitter and base-emitter capacitances, is this reasonable?
Yes, since we are dealing with audio frequencies and low impedances it is completely reasonable.
- Did you make any mistakes in the execution?
Well one thing that I will note is that I would definitely not, ever, have connected the gnd of my oscilloscope to anything other than gnd, Vcc or eq. (low impedance point) of my circuit. This is because inside the osc. its gnd is capacitively coupled to earth, or worse, if not earthed it can even be capacitively coupled to line, through a mains input filter.
Another thing is related to the circuit diagram that you show in your question;
The FET used to switch the collector current is an NMOS, and it has its source connected to the supply, I assume that this is an error, and that you meant for it to be a PMOS?
Also, I did not address the point of how you drive the FET, because you did not show any gate-drive circuit, and I have just assumed that you have measured the collector voltage, and that it does indeed switch fully on and off.