I think I understand your confusion.
You are trying to power a fridge, don't you?
The VA measure ("apparent" power) has only indirect relation (see the p.s.) to the efficiency. It is a property specific to the AC power setups (not only inverters at any rate).
At any given moment, the power transferred is a product of the current and the voltage in the circuit.
The AC voltage switches direction many times per second (depending on where you live, it is either 50 or 60 cycles per second, this is what the "Hz" measure is for). The AC current does the same, but depending on the load type, it may lag or advance in relation to the voltage. This amounts to the load returning some part of the energy back to the source in each cycle.
If the load misbehaves like this, the "real" (averaged) power transferred from the AC source to the load will be less than the "apparent" power (the product of the averaged current and the averaged voltage).
The ratio between these two "powers" is called a power factor. Good AC loads have a high power factor (near 1.00 or 100%), bad loads have less.
Examples of "good" loads in regard to power factors are incandescent bulbs, heating appliances and modern electronics.
Examples of "bad" loads are most types of electric motors, cheap LED lightbulbs and older electronics.
A sane and efficient inverter is expected to consume input power related to the "real" output power (W) and not to the "apparent" output power (VA). In your case, it could be something like 200W (allowing for ~90% inverter efficiency, normal for a modern inverter).
On the other hand, the inverter output stages need to be engineered for the "apparent" power that may be higher than the "real" power of the load.
This is why inverters have both "real" power (W) and "apparent" power (VA) ratings and this is why your inverter reports both values.
If you keep adding load to your inverter, you could overload it both by the "real" power consumed by the load and by the "apparent" power, independently from each other.
p.s. A minor (in your case) consideration could be that a load with a bad power factor somewhat lowers the inverter efficiency. But this effect is quite minor for a modern inverter and in your case maybe amounts to additional 5W at the inverter input.
p.s. #2 If you want to estimate the real efficiency of your inverter, you need an additional value - the power consumed at the inverter input. If the inverter does not report it (most of them don't), you need a device that measures the battery voltage and the battery discharge current. Devices like this do exist, but they don't add much value to setup like yours so they are rarely used.
p.s. #3 in regard to your comment about the awful efficiency, ~60% is quite low and neither to be expected from a brand new inverter nor particularly safe for it. This would mean that those missing 100W become heat at the inverter. This is a lot and the fan would run almost constantly. One would expect like 2kW load in order to get 100W loss. At least this is how a similar inverter of mine behaves (24V input, 2kW/4kW peak). It has label efficiency of 92-97% and is almost the cheapest I found.