The temperature of the circuit can probably be crudely modeled by assuming that you have two channels: one for heat absorption, and one for heat loss.
Heat loss
Suppose you have something like a light bulb.
The hot filament loses heat via radiation and conduction.
Let's focus on radiation.
The rate of heat radiated by an object of temperature $T$ depends strongly on the materials in the object.
In general, the Stefan-Boltzmann law says that the total power radiated by an object of surface area $A$ and emissivity $\epsilon$, and temperature $T$ is
$$P_{\text{radiated}} = A \epsilon \sigma T^4 $$
where $\sigma$ is a the Stefan-Boltzmann constant.
The emissivity $\epsilon$ is highly material dependent.
Let us just suppose it has some unknown value for the light bulb filament and combine all the constants in the Stefan-Boltzmann equation in to a constant $C$, giving us
$$P_{\text{radiated}} = C T^4 \, . $$
Heat absorption
Resistance in the light bulb filament happens roughly because electrons collide with the atoms in the metal, causing them to vibrate.
This process, called Joule heating, causes the metal to absorb power given roughly by
$$P_{\text{absorbed}} = I^2R \, . $$
Heating rates
Now we have two rates, $P_{\text{radiated}}$ and $P_{\text{absorbed}}$.
Assuming that the temperature of the object is proportional to its internal energy
$$T \propto E$$
and recalling that power is defined as the rate of energy change, we get
\begin{align}
\frac{dT}{dt} \propto \frac{dE}{dt} &= P_{\text{absorbed}} - P_{\text{radiated}} \\
&= I^2 R - C T^4 \\
&= I^2 R_0(1 + \alpha (T - T_0)) - C T^4 \, .
\end{align}
This says that when you turn on the light bulb (in this case with a constant current source), the temperature is time dependent.
This means that the resistance and therefore the voltage are also time dependent.
However, note the important fact that the right hand side goes to zero for a certain value of $T$.
This means that the bulb will heat up as time passes, but will eventually reach an equilibrium temperature $T^*$.
We can therefore assume that there exists a function $g$ such that $T^* = g(I)$.
At this point note that if we have a constant voltage source instead of a constant current source we would have had
$$ \frac{dT}{dt} \propto \frac{V}{R_0(1 + \alpha (T - T_0))} - CT^4 $$
but the point about having an equlibrium temperature and a function $T^* = g(V)$ would still hold.
Comments on the equations in OP
The point is that your equation
$$R = R_0(1 + \alpha g(P))$$
should probably be more like
$$R = R_0(1 + \alpha g(I)) \qquad \text{or} \qquad R = R_0(1 + \alpha g(V)) $$
and this only holds when the system is in equilibrium.
For a light bulb I'd guess that equilibrium is reached within a few seconds, but I don't really know for sure.
Anyway, in the equilibrium case $V$ is some implicit function of $I$ (or vice versa for a constant voltage source) so you can still write
$$\text{unknown function}(IV) = \frac{1}{\alpha} \left( V/I - R_0 \right)$$
as you already did.
