The power output of the water heaters is the crucial point. The "high performance" heater is 76 kBTU/Hr, while most water heaters only are around 30-40 kBTU/hour. This larger burner allowed the smaller tank to be able to "keep up with the demand" for quite a while. Another reason is because it maybe would be configured to heat to a much higher temperature, and then use the mixing valve that in includes to reduce the temperature to the reasonable level. I'd consider this "cheating" if it is what they are doing, though it is allowed by the government.
I'm guessing that the listing says one person because it is a small tank, and not based on the FHR (which they should have paid more attention to).
I don't think that you are missing anything based on the specs on FHR. But, you do need to pay attention to the size of your natural gas line. You'll need a higher flow rate with this model since it's usage rate is much higher. You also might need to provide more "leaks" in your house (and room containing the water heater) to allow for it to get sufficient air for combustion. This particular model uses air from its surroundings, instead of having a pipe to bring air in from the outside. Insufficient fresh air will generate CO (very bad for living things) and will decrease the water heater's efficiency. Make sure that you have a CO detector installed near the unit, whatever your choice ends up being.
The first hour rate is an odd metric. It measures the number of gallons that can be supplied at above 110 F, with an initial temperature of 135 F. This is very difficult to use in calculations.
I'd approach the calculations as an energy conservation problem. Use the recovery rate to calculate the burner output (1080 kJ/min in this case). Then (ignoring that entropy is gained in the mixing valve and the heat capacity and density of water depends on temperature), decide on what temperature shower you want, and what temperature water you want in the tank. Use the heat capacity of water to calculate the amount of energy stored in the water heater, the energy usage rate (power out), and the burner energy rate (power in). Assuming use of 6 gpm, input water at 7 C, the tank at 75 C, and using water at 38 C, I calculate that you would have about eight minutes of water. With the 55 gallon tank, you'd have about 18 minutes using these assumptions.
Just by looking at the 90F recovery rate, you get about 80 gallons/hour. So, you could use 80/60=1.33 gpm indefinitely (with a rise of 90 F). But, you probably only need a rise from 45 to 100 F for your shower. In this case (55 F rise), you could use about 2.2 gpm indefinitely with this particular burner.
It's mostly the gas pipe. A tankless water heater is going to need a 1" or even 1.5" pipe to feed it. Your current water heater probably uses a 1/2" pipe. The new pipe probably needs to come off the main trunk, and replacing it requires a precise gas volume calculation and gas plumbing work that is very skilled labor and not cheap. If your plumber can't do this work himself, he's probably subbing it out and you're eating the cost.
Secondarily, there's also the re-plumbing of the water lines to feed the new heater. The inlet and outlet ports are probably in different places compared to where they are on your tank water heater. By contrast, if you replace it with another tank water heater, the water lines are already in the right place, and all you need to do is connect the tank with flex lines.
Finally, the exhaust may be different, too. Particularly if you go with a high-efficiency model, it can't be exhausted through the current pipe that's likely going through the roof.
Best Answer
Model the water heater as a continuously stirred tank reactor (CSTR), so it is always at a uniform temperature. Assume the recovery time is not dependent on temperature and completely accounts for insulation losses and the like. Neglect losses in pipes and assume the operator controls the shower temperature to 105°F perfectly. Taking the stopping criterion from the question, the shower is over when the water in the tank becomes 105°F.
The recovery rate raises 140 gallons of water by 81.3°F in one hour. We'll say this is a constant heat input of 9118 W.
The key is that a constant-temperature shower removes a constant rate of heat from the tank, that which is associated with raising 2.5 gpm of water from 58.7°F to 105°F. This is 16986.5 W.
The difference is 7868.5 W. With constant heat, you don't need any complicated integrals.
The time for the tank to drop from 140°F to 105°F is 1568 s or 26 minutes.
The (maybe counterintuitive) fact that the variable flow rate from the water heater does not influence the rate of heat removal from the water heater comes from the fact that the incoming cold water is the same temperature at the shower and at the water heater.
Note that, because
does not appear in the expression for 
, 
is now a constant.