Impedance matching and power transfer
Am I missing something?

No, you are not missing anything. If you design an impedance matching network (L-Pad, pi-network etc.) then it is lossless.

How is the power balance in that case?

If you are matching a resistive source to a different resistive load, the powers are still balanced as per this example: -

Image from this calculator.
In the above example you'll see that it calculates the voltage gain as 2.44949 so, if 1 volt is fed at the input, the output will be 2.44949 volts. If you calculated the load power and input power (1 volt into 50 Ω) it equals the output power.
Of course there is still the same power dissipated in the source impedance but it sounds  like you realize that.
If your load is complex then make it resistive with the appropriate conjugate component. Ditto the source impedance or, both.