Electrical – What voltage are AGM battery capacity ratings supposed to be computed at

Your 75W fan required 220 Volts, it will not operate at typical battery voltages such as 12V.

A 75W fan operating (indirectly) from a 12V supply would draw 6.25 A not 0.34 A.

Even with a theoretical 100% efficient invertor, the current on the 12V side is greater than the current on the 220V side to deliver the same power across the system from battery to Fan.

Note that the reason that power is delivered through the grid at high voltage is to keep currents low. losses and cable sizes are proportional to (a function of) current. High V are used to keep currents low.

You convert 12V DC into 220 V AC using your inverter. Depending on capacity and utilisation the efficiency of an inverter might be the 80% you estimate or it may be worse - 50% Ref So your 6.25A may become 13A drawn from the 12V battery.

Therefore your 300 Ah battery-set might power a 75W fan for less than 24 hours.

Update: for 2 x nominal 12V SLA battery, the following diagram may clarify what is happening.

^{simulate this circuit – Schematic created using CircuitLab}

A battery has some internal resistance that for Li-ion batteries is usually in the 0.15 \$\Omega\$ region when fresh.

As you should know from Ohm's Law, \$V = I \times R\$ so if you put a load on the battery this will cause a current to drain from the battery and this will cause, depending on the current, a small voltage drop over the battery's internal resistance as it would on any normal resistor in a circuit. This can be explained visually like so:

^{simulate this circuit – Schematic created using CircuitLab}

As the battery gets used and 'ages' this internal resistance increases so the voltage drop over it will also increase for a constant current. The 'open-circuit' voltage of the battery will always be more or less the initial voltage as the 'load' will be near infinity so all the voltage drop will be over this air-gap between the batteries two terminals.
As you mentioned the drop in voltage is less with a larger load, this is because the current draw will be less and hence the drop over the internal resistance will also be less.

You should look into a constant current source using a low voltage op-amp (I did something similar to this a couple of months a go and ran into the same issue - hence that question) and you'll be able to work out the mAh easily enough as the current should be more or less constant and you'll know the time taken for the voltage to fall.

Hopefully this helps and answers your question - it is expected that a voltage drop will be seen when applying a load to a battery.