Does Impedance Matching Imply any Practical RF Transmitter Must Waste >=50% of Energy?
If your power source is a zero ohm output voltage source, followed by a 50 ohm resistor, then yes, what you think is correct.
However, practical RF amplifiers (at least ones designed to be efficient) are never built like that. They tend to have a low impedance common emitter or source stage followed by reactive impedance matching, all designed to operate into 50 ohms.
Interestingly, if you buy a general purpose signal generator, the output is usually built as a voltage source followed with a real 50 ohm resistor, as efficiency is not an issue, and having an accurately defined output impedance over a very wide frequency range is the main design goal.
RF amplifiers do NOT in general have an output impedance remotely close to 50R..... They are however designed to drive a 50R load!
Much like audio amplifiers the source impedance is generally far from the design load impedance, because you DONT want maximum power transfer, you want something closer to maximum efficiency!
Depending on the topology, the things approximate either voltage sources (Low output impedance) or current sources (High output impedance).
If you think about a for example, HF, Push pull output stage, the devices are operating at some designed voltage and current, hence some 'impedance' (Usually quite low), which is then transformed to an industry standard 50R.
This impedance is set by the designer to result in some voltage across that 50R load that will deliver whatever the designed power level is. Note that those output devices could be in deep class C or even class F and be operating essentially as switches dissipating near zero power, but I as a designer still need to decide what voltage and what current to chose as an operating point and hence what transformation I need to get to the target power at the output.
Now clearly if you try to run such an amplifier into a load far removed from 50R then the voltages and currents seen by the power devices will be other then the designer intended, and if you go to far the smoke comes out.
A further complication is the output filters and (at UHF and up) the possibility of a terminated circulator at the output which actually makes the thing look like 50R looking back into the input.
So, is it correct to say impedance matching implies the efficiency of any practical RF transmitter cannot be greater than 50%? And any practical RF transmitter must waste at least 50% of energy?
No, that's wrong. The diagram in your post lacks the essential building block in this discussion: the amplifier itself.
All amplifiers can be described according to their PAE (Power Added Efficiency).
$$ PAE = \frac{P_{out}-P_{in}}{P_{supply}} = \frac{P_{out}-\frac{P_{out}}{G}}{P_{supply}} = \frac{P_{out}}{P_{supply}}\left({1-\frac{1}{G}}\right) = \eta\left({1-\frac{1}{G}}\right) $$
PAE is the key parameter here, because the amplifier gain is likely to be very high. The power transferred TO THE AMPLIFIER by the generator, when impedances are matched, will be only 50% of the maximum generator power indeed. But if gain is high enough then the power wasted in the internal impedance of the generator will be very low compared with the power delivered BY THE AMPLIFIER to the load. Thus, the impact in total efficiency is likely to be low.
What matters here is (mainly) the output stage of the amplifier having a high raw efficiency \$\eta = P_{out}/P_{supply}\$, which depends on the amplification class (A, B, AB, C, D, F, etc.) and the operating point of the amplifier.