Does the source impedance of DC-AC converter always represent losses?
It depends what power you choose to deliver from your source.

If you choose to deliver the maximum possible by varying the load impedance to be equal to the source output impedance, then yes, your efficiency is limited to 50%.

If instead you choose to deliver much less than this maximum power, your efficiency can approach 100%.

Consider a 200v source, with a 0.1ohm output impedance. This might be a reasonable model of a mid-sized DC-AC converter.

If you choose to try to extract maximum possible power by running it into a 0.1ohm load, then 200v into 0.2ohms gives you a load current of 1000A, with 100v into the load, delivering 100kW at 50% efficiency. 

However, I did say this was a mid-sized converter, perhaps 2kW rated power, it's not going to handle 1000A output current, even though it does have a low output impedance. At 2kW, 200v, we would be running 10A load current. 10A in the output impedance of 0.1 ohm would drop 1v, so the efficiency due to the output impedance alone is around than 99.5%.

Just because a converter has a low output impedance doesn't mean it's capable of delivering the matched load maximum power output.