The boiling water is converting liquid water to gas. Unless this gas is continually removed by the pump, it quickly increases the pressure inside the vessel. This increased pressure will stop the boiling.
Setting a lid on the jar gives it a one-way valve. Gas can still escape. If you instead put on a full seal so that gas cannot escape, then it will stop boiling even with the vacuum maintained in the outer chamber.
Firstly, it would be better to use actual, accurately measured numbers than the human 'experience': the human body is a poor thermometer and the mind plays tricks on us.
But that hot water droplets lose heat and thus cool down in cooler air is an established fact and a consequence of the laws of thermodynamics.
Regarding your three first bullet points, despite some limitations you point out, those do not mean a hot droplet of water doesn't cool in air: it does and partly in accordance Newton's cooling law.
As regards work done by the droplet (overcoming the viscous drag), if anything that would lead to heat generation, not cooling (but the effect is truly minuscule).
Kinetic or potential energy of the droplet have no effect on the droplet's temperature. Temperature is simply a measure of the average speed of the molecules of the water and that is not affected by these energies. Spinning water in an ultra-centrifuge does not make its temperature rise, for instance.
You have however overlooked one major cause of heat loss: evaporation. Your shower 'steams up' because hot water evaporates and that costs energy, known as the Enthalpy of vaporisation.
Millions of tons of water are cooled this way everyday in power plants world wide: the cooling towers drop hottish water from the top of the towers and evaporative heat cools down the water (the evaporated water escapes as steam clouds).
If your shower has been in operation for a long time and the bathroom's temperature is equal to the shower head's water temperature and the air is saturated with water vapour, then no cooling of the shower water would take place.
Best Answer
If heat were being supplied to the boiling water to hold its temperature constant at room temperature, then the pressure of the water vapor in the gas phase would be equal to the equilibrium vapor pressure of water at that temperature. This would be roughly 17.5 torr. If you then stopped the boiling, the pressure in the gas phase would remain 17.5 torr. Now, if you introduced air at room temperature into the gas space at 1 atm. (760 torr) and mixed the gas very rapidly, there would be no change in the partial pressure of water vapor (provided the size of the container did not change). It would remain at 17.5 torr, while the air pressure would adjust to 760 - 17.5 = 743.5 torr. Essentially none of the water vapor would condense.