Yes, that's why I said moving the air takes "a totally insignificant amount of power
compared to the actual refrigeration involved." You're talking about the actual refrigeration.
The terminology you need to google for "the heat that is created when the water in the air turns into a liquid" is "enthalpy of vaporization of water"; around room temperature this is about 2.4 MJ/kg, so 10 liters/day is 24 MJ/day, which is 6.8 kWh/day, or, in SI units, 280 watts. However, remember that a heat engine operating at the Carnot limit is reversible. The coefficient of performance of a typical heat pump at these temperatures, the kind you might buy off the shelf at a big-box hardware store, is about 2, so you only need about 12 MJ/day (3.4 kWh/day, 140 watts). At a typical desert capacity factor of 25% this means you need a 560-watt solar array, about US$120 and three square meters. (California's utility-scale PV average capacity factor was 29% last I looked.) Very cloudy and polar places can have PV capacity factors as low as 10%, but they also have easier sources of drinking water. Like a rain barrel.
It's easier to store water or to "store coldness" than to store electricity, so you don't need electrical storage, you just need a heat pump sized for your peak throughput instead of your average throughput. Heat pumps are pretty expensive, so you might think this is a big problem, but the cheapest air conditioners I can find for sale around here are about 2000 watts, not 500 watts.
I've been noodling on desiccant-powered heat pumps for this and other uses, which may be able to reduce the cost of such systems, gather a larger fraction of solar energy than the 21% of high-efficiency PV panels, and provide built-in energy storage.