Your calcs are OK except for the second case. The claim is up to 140F, which translates to 60C. You started by assuming the initial temp is 30C so that you had to heat by 70C. But if your target is 60C, not 100, then the temp rise is 30C, not 70, and the heating time calc comes to 4.13 hours (not just over 2).
Hugest problem is your calcs is heat loss from the heating cup. You hypothesize "ideal case" by which you appear to include NO heat loss by perfect insulation. Ain't gonna happen, ever! Even with good insulation you would surely reach a temperature of the mug and coffee that creates a heat loss of 2.5 watts exactly offsetting the heat input, and temp would never rise further. Moreover, because of this real heat loss rate (which depends on the cup temp), the time to get up to that equilibrium temp would be MUCH longer than your 9 hour estimate. By the way, my suspicion is that, on the website you linked for the commercial heating pad, what they did not tell you is that the device may be able to KEEP at mug of coffee at 140F, but only if it is put on the heating pad already at that temperature. In other words, they estimate that the heating input rate their unit provides will just manage to match the heat loss rate from the mug at 120 to 140F.
Another big problem with your calcs is you ignored the mug. To heat the 296 g of water (coffee, whatever), you also will have to heat the mug itself, unless you can produce a perfect mug that never steals heat from its contents. The mass of the mug probably is similar to the mass of the contents. If we knew the heat capacity of the mug material, we could calculate its heat requirements to keep up with the warming coffee, and that will be in the same ballpark. So now your time estimate is roughly doubled, but not known precisely.
By the way, you also made a slight error in stating the problem. You said, "Boiling water via USB port". What you actually calculated was how to get the liquid water temperature from 30C to 100C. That is, the water is as hot as boiling water, but it has not actually been boiled to convert it to water vapor. And for purposes of drinking hot coffee, that is just fine. But to BOIL water means to finish the conversion of it to water vapor, and that's a huge surprise to many people. The process of converting liquid water at 100C to water vapor still at 100C requires more heat input for the change of state. The process is called vaporization (no surprise there!), and the Heat of Vaporization of water is 100 times the Heat Capacity of water! It's a strange coincidence, actually. With a Heat capacity of 4.186 J per gram-degree, it takes 418.6 Joules to raise the temperature of one gram of liquid water from its freeing point, 0C, to its boiling point, 100C. But then to convert that same gram of water liquid to vapor takes an additional 418.6 Joules of energy!
Oh, and the comments about 100% efficiency are a little misleading. Efficiency usually is the word to describe how much of the energy input (electrical from the USB port) actually gets converted into usable heat. IF the heating device you postulate is a simple resistor, it is guaranteed to be 100% efficient. ALL of the energy flowing through a resistor is converted to heat, and none gets lost, destroyed, or stored. On the other hand, once that conversion to heat is done, all the other processes concerning where the heat goes and how much gets lost from the coffee to the surroundings have a big impact, as all have said. If you include those processes in the concept of "Efficiency" of the system, then it certainly is not "100% Efficient".