Do you not see the similarity - the direct correspondence, in fact?
I thought we'd already agreed that, after a long time, nearly every charge can be thought of as getting all the way through the resistor. The ones that do not have a complete journey (after switch on or just before switch off) will only have a portion of the PD to 'drop through'. But, the same number of charges make incomplete journeys at the start and the end of the experiment and the total Potential energy of those charges converted will be the same as for one 'resistor-full' of charges.
Why not do a series of sketches showing charges at various points along the resistor and at different potentials?
Imagine I have a 1kg mass and I let it fall distance s, working some machine. I take a different 1kg mass and let it fall from the end point of the first mass's journey a further distance s. The two masses will have converted 2sgJ of energy - just the same as one mass, falling through 2s. If this were water in a hydroelectric dam and the valve was opened, the water at the bottom wouldn't supply much energy before exiting the system and neither would the water that only gets a few metres down from the top before the valve is closed. But, altogether, we are always dealing with a pipe full of water when we want to work out the energy obtained. Likewise, we are always dealing with a Resistor Full of charge when calculating the Power dissipated when the circuit is connected. The speed that the charges move is totally irrelevant in that respect.
I don't think I can help you any more with this. I've put it in as many ways as I can think of. It's up to you to convince yourself now. The theory is well established as correct and the models are valid.