Your approach is basically correct, but I question some of your numbers:
How does "1 day of reserve" only require 30% additional capacity, rather than 100%?
You've neglected to take into account the storage efficiency of the battery itself (Wh in does not equal Wh out), which is only about 70% for lead-acid.
For your 275 mA × 9 h = 2.5 A-h daily load, you need a 10 A-h battery, which accounts for your 100% functional reserve, plus 100% "battery reserve".
Your charger will need to supply 7 A-h per day, which is the functional capacity (including reserve) divided by the storage efficiency of the battery.
If you have a very good charge controller, it might achieve 90% power efficiency (but possibly much less). 7 A-h × 12 V = 84 W-h nominal.
84 W-h / 0.90 efficiency / 4 h (equivalent insolation) = 24 W panel.
First, the annual solar irradiance is calculated by taking the power incident on a square meter when perpendicular to the sun, and multiplying by 24 hours and then by 365 days. Since the irradiance is typically about 1.366 kW/sq m, when you multiply it out you get 11.9 kW-hr/sq m, which is your number.
But you might have noticed a problem. This assumes the sun is shining 24 hours per day, which is hardly proper.
Second, your second calculation also assumes 24 hours per day sunlight, which is not right. Since the average length of day is about 12 hours, your 25.6 Wh should be divided by 2, giving 12.8 Wh per day.
Finally, this number tells you how much power is falling on the solar cell, not how much the cell puts out. It also assumes the cell will be perpendicular to the sun at all times, and that the atmosphere does not attenuate sunlight. The first may be true, the second is not. Consider that the sun is much dimmer at sunset than at noon.
Let's take the case of noon sunlight. Assuming the air is very clear and you are getting the nominal irradiance, the power falling on the cell will be 1366 w/sq m x .0078, or about 10 watts. Since you have a solar cell rated for 1 watt, this says that you can assume the cell has an efficiency of about 10%, which is about right.
For a 1-watt solar cell which is fixed in position, and is perpendicular to the sun at noon, you can figure on a total output of 4-6 Wh on a clear day. On the one hand, as the sun moves away from the noon position, the effective area decreases (reaching zero at sunrise and sunset when the cell is edge-on to the sun). Also, sunlight gets attenuated by the atmosphere at lower angles, and finally the cells themselves will typically become less efficient at lower intensities.
Best Answer
I assume your system "consumes" 0.5 Ampere of current and not 0.5 Amperehours of charge (that would be an unusual measure).
So you have calculated the power of your system to 5 V * 0.5 A = 2.5 W. It will run for 48 hours, so it needs 2.5 W * 48 h = 120 Wh.
So you'd need a battery with 120 Wh, but there is no ideal 5 V battery and no ideal 5 V solar panel.
You chose (I suppose) a 12 V battery and a solar panel with charger for that. And you want to charge the battery in one day so your system can run for two days without recharging.
In an ideal world you need a solar panel of 120 Wh / 6 h = 20 W (that is 12 V with a current rating of 20 W / 12 V = 1.67 A). And a battery with 120 Wh / 12 V = 10 Ah.
But the world is far from ideal:
To get from the 12 V of the battery to 5 V of your system, you'll use a switchmode power supply. Let's say you get a not very well build one (because it's cheap) and it has only an efficiency of 80%. The battery must have more energy to power the losses of the power supply. So with 80% you would need 120 Wh/80% = 150 Wh.
Next thing to consider is that for improved battery life (if that is an issue) you don't want to have it cycle from 0% to 100% often, but more like only from 25% to 90% (or even less). So you only use 65% of the rated capacity. For 150 Wh needed energy you'd use a 230 Wh battery.
The increased need alone now requires the solar panel to deliver at least 25 W. But sizing the solar panel based on peak power and sun hours is asking for trouble (except for some very sunny regions I guess). So to get a better estimate, you look up some statistics of the area where it is supposed to be used. At my place you have a measly 1.5 sun hours in December, and a day is roughly 8.3 hours long. Of course your solar panel will produce some power even when you don't have direct sunlight hitting it, but it's far from the peak power. So maybe it's 100% peak power and 30% power (I made that number up, no idea how much it is) for the rest of the day. So you'd get 25 W * 1.5 h + 30% * 25 W * (8.3-1.5)h = 85 Wh. We need roughly twice that amount. So better go for a 50 W panel.
I haven't even mentioned that you need a maximum power point tracking charger to get that, so the charger will have a certain efficiency which reduces the amount of energy available to charge your battery, think of another 90% efficiency and you need 55 W.
There should be better estimates around on how much solar energy is available. Like this graph:
(taken from Wikipedia by SechWatt)
It shows the total energy produces by a 1 kWp (kilowatt peak) solar panel per month somewhere in northern Germany. The average day (from sunrise to sunset) in December is 7.8 hours there, so you have 20 kWh in December, which averages to 20 kWh / 31 days = 645 Wh per day for a 1 kWp panel. With our needed 150 Wh we end up with a (150 Wh)/(645 Wh) * 1kWp = 232 Wp solar panel. So my estimation of 30% was probably way off.
Note: There should be calculators around for this kind of analysis.
If you plan to use the system for several years without replacing the components, you have to factor ageing in as well (battery capacity reduces, solar panel power reduces). So that makes things even larger.
Conclusion:
Use a battery rated for 230 Wh (12 V / 19.2 Ah), and a 232 Wp solar panel (12 V / 19.3 A), if you want your system to work in December in northern Germany.
If you plan to use it elsewhere, calculate again.
This should only be considered a rough guideline on what should be considered and I wouldn't consider it a complete analysis.