This problem is very simple, but it's easy to overcomplicate by looking at too small a scale. At every second – no matter what the heater does – you waste money by heating the outside of your house. The rate of heating – and thus the rate at which you waste money – is given by Newton's Law of Cooling. So
$$ \text{Wasted money} \propto \int (T_\text{in}-T_\text{out})\;dt $$
The lower your house's temperature, the less money you waste – no matter what. So set the thermostat to the lowest practical temperature when you're away.
There are too many unknowns to model the situation with any accuracy.
If you know the power being pumped into your pan then you can easily calculate the amount of steam generated from the latent heat of vaporisation of water. You can calculate the power being generated by your cooker from the flow rate of the gas, or the current if it's an electric cooker, but it's anyone's guess what percentage of this is lost to the environment and how much ends up in the pan.
For any significant flow rate of steam I would guess that turbulent mixing will ensure the steam temperature is roughly constant throughout the pan. The temperature will be whatever the boiling point of water is at the internal pressure. The steam flow rate through the hole in the pan is fairly straightforward to calculate from first principles, though since steam is so important industrially I'd guess some Googling will find tables and empirical equations for flow rate.
Response to comment:
Let the mass of water lost per second be $m$, then the power applied to the water in the pan is just:
$$ W = mL $$
where $L$ is the latent heat of vaporisation of water. This will be equal to the power generated by your cooker times some unknown factor less than unity to allow for heat loss to the environment.
Response to second comment:
The temperature of the water will be close to the boiling point because any water hotter than the boiling point turns to steam, and the latent heat required will cool the water again. The steam in the layer immediately above the water will be at the same temperature as the water because it's in thermal contact with it.
If the steam above the water is hotter than the water you have to ask what is heating it. The only things I can think of that could heat the steam are the pan walls and lid. However the pan is only being heated from the bottom, and heat flow by conduction though the pan walls is a lot slower than heating/cooling by convection between the pan walls and the water in the pan. Therefore I would guess that the pan walls and lid are also close to the boiling point of water - actually they will probably be slightly cooler because they will lose heat to the surrounding air.
So I would guess that as long as there is enough water in the pan the steam in the pan will be close to the temperature of the water.
Best Answer
If you are using a good temperature controller it should hold that temperature at the set point almost regardless of external atmospheric temperature. As the external temperature rises it would use less power to maintain the fixed set point. As external temperature fall it will compensate by adding more power to match the heat loss from the tank.