unless the battery management/charger IC specifically does it, your system load WILL drain your battery to death. I have that issue on my BQ24090 single cell lipo charger i put on my little combat robot. The output of that charger goes to the lipo, but that it also the system load rail!
You actually need to monitor this issue yourself in most cases: one nice way to do it is monitor the battery voltage using a micro controller (your arduino!) ADC or even an opamp (probably better, then it's not relying on software to do it) which sees that the Lipo is now at 3V (safer to stop here, dont go so low as 2.75V) and then cut off/turn off the system. At this point you assume "the battery is flat" and you deal with it - in your case you would disconnect the load (system, not the lipo) until the battery voltage has been beefed up enough by your solar charge system to turn on the system again. Maybe the "turn back on" voltage is 4V?
do some investigations into battery monitoring/protection ICs, but basically you need a more complicated one to recover after the solar cells have brought up the battery level enough to continue again. Otherwise you can assume that a human can just replace the battery, and deactivate the "load disconnect" switch, and let the system continue now on a full battery?
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.
Best Answer
The documentation states that until the battery voltage reaches 3V, it will only perform trickle charging of the battery (very slow charging), at a maximum of 90 mA. This is to protect the battery:
Once the battery reaches 3V, it will start constant current charge at up to 900 mA, until it reaches the 4.2V cutoff, at which point it switches to constant voltage charge. So it could take a little while for the battery to get back to a state where you can actually draw power from it.
There's apparently a minimum battery voltage of 2.4V, under which output from the battery is cut. If your battery is discharged, with a load connected its output may drop below that even if your battery has been charged a little bit.
Just leave the solar panel connected longer. After a slow charge period until it reaches a safe level, it will start faster charging. Once charged enough you should be back to the initial state.