Your question initially seems to focus on whether a counter process exists to heat a cold liquid that is the inverse of the process of natural convection + evaporation that cools liquids.
The short answer is ... No, such an inverse process does not exist entirely as stated.
First, the presumption we will make is that the liquid is below its boiling temperature at the given pressure. Any heat transfer to the liquid only raises the temperature of the liquid, it does not cause the liquid to boil.
Let's take the case when the air has no vapor for the liquid component. This addresses evaporation by making it present in both cases, hot or cold fluid. Let's also set the air temperature as a constant $T_a$. This allows a direct comparison by vapor pressures only of the liquid.
An open mug (open at the top) with a liquid at $T_h$ that is hotter than the surrounding air at $T_a$ will experience natural convection and evaporation to the surrounding air.
An open mug (open at the top) with a liquid at $T_c$ that is colder than the surrounding air at $T_a$ will not experience natural convection processes. Since the air has no component of the liquid, the liquid will still evaporate. Since the vapor pressure of the cold fluid is lower than that for the hot fluid. The evaporation rate of the cold liquid to hot air will therefore be lower than that for a hot liquid to cold air.
In conclusion, all else being equal for the same temperature difference ($T_h - T_a$ = $T_a - T_c$) for a liquid in an open mug, a hot liquid will cool faster than a cold liquid will heat up. Natural convection occurs in the case for the hot liquid. The evaporation rate is faster for the hot liquid.
The relative ratio of heating/cooling will depend on the temperature difference ($T_h - T_a$ versus $T_a - T_c$), the relative saturation of the air with the liquid component (essentially a vapor pressure to partial pressure difference), and the absolute temperature of the liquid (to set the magnitude of the vapor pressure of the liquid itself).
Also, when both mugs are closed off entirely and all else is exactly equal, both systems will have the same rate of temperature change. All else in this case is emissivity (as it affects radiation to/from the mug and surrounding air), thermal conductivity (as it affects conduction through the mug), and convection coefficient (as it affects convection to/from the mug and surrounding air).
Finally, when the liquid is at its boiling temperature for the given pressure, heat transfer into the liquid will cause the liquid to boil. In an open container, a film of cooler vapor will form above the liquid as a boundary to the hotter gas (air) in the surroundings. This will impeded the conduction of heat from the hotter air back to the liquid. Heat instead will go first from the surroundings to heat up the vapor above the liquid, and then the vapor will rise out of the mug. So, all else being equal, a boiling liquid can in practice experience lower heat transfer rates from the gas above it.
The comment added clarifies the question further: Does a cold liquid heat up faster when it is a covered mug or an uncovered (at the top) mug.
Fill the mug to the brim or to the top (no extra air space). Set the air temperature to be above the liquid temperature and keep it stagnate (not flowing). Allow that we are talking about heat flow through a thermal conductor (not a thermal insulator) in the direction from the (hotter) surroundings to the (colder) liquid. Normalize the heating rate to the rate of heat transfer per unit volume of liquid for both cases.
Opening the top of the mug changes heat transfer modes from the top only.
In a closed mug, heat conducts through the top cap into the water. In an open mug, heat flows from the hot air to the cold liquid through a boundary layer vapor film above the cold liquid. Conduction rates of heat through a gas boundary layer are lower than conduction rates through the same thickness of a (thermally conductive) solid. As the solid cap is thicker, the conduction rate across it becomes smaller. As an order of magnitude, the thermal conductivity of stagnate air is perhaps a factor of 10x lower than glass. For a 100 micron air boundary layer above the liquid, a 10 micron ceramic plate has the same conduction rate, and a 1 mm thick ceramic plate as 100x smaller conduction rate.
--> The jury is out here. When the solid cap is thicker than the boundary layer using the ratio of thermal conductivities of the air to the cap material, the open mug will have more heat conduction from the top air.
Evaporation occurs from the open mug (and not at all from the closed mug). The evaporating liquid builds a colder vapor layer above the liquid. This slows down the conduction rate to the liquid below because heat that would go to heat the liquid is instead taken to heat the cold vapor.
Finally, in the open mug, the surrounding gas radiates heat to the colder liquid. In the case that the cap is only as thick as the boundary layer on the open mug and when we neglect evaporation as a process to limit heat transfer to the open mug liquid, radiation can be a deciding factor to make the open mug heat faster.
The jury is out in practice. You can set the restrictions on the system configuration, apply first principles, and reach a conclusion either way. Otherwise, the arguments become "what-if" cases that go on forever.