Ha, ha, nailed it!

I mean, cobbled together a schematic that is simple, yet very sturdy against loading conditions, and high speed, too.

The main hint was this:
[Current-Feedback] ... CFB amplifier, which is typical as AWG output amp ...
That made me curious to learn more about Current-Feedback (CF), and digging through literature found that multiple references were pointing to a Sergio Franco's article in the EDN magazine from Jan 5, 1989: "Current-feedback amplifiers benefit high-speed designs", which happens to be available at the archive.org, as a scan of the paper-printed magazine (see page 161 of 286):
https://archive.org/details/edn-1989_01_05/page/160/Once knowing the basics, it was an exercise to convert this:

which uses voltage-feedback and increases only the max current, into a current-feedback amplifier. Also wanted to increase both the max current and the max voltage of the opamp, so this is what I came with:

To understand why that works, let's remove the current feedback resistor, Rcf.
The idea was to use the opamp as a voltage repeater, so the output of the opamp to become a zero-impedance inverting-input for the CF network.
Since the positive-input of the opamp is tied to ground, and with the opamp as a voltage repeater, the output of the opamp will always be zero, so no need for a load resistor for the opamp. The output of the opamp (which becomes the inverting input for the CF network) will appear as if it were acting on a zero ohms load (a zero ohm virtual load).
The zero ohms load is virtual, but the current has to go somewhere, so the current will close through the power supply pins of the opamp. This means the current through the emiters of Q1/Q2 will mirror the virtual current through the zero-ohms opamp's virtual-load.
R1 and R2, together with the quiescent current of the opamp will set the minimum idle current through Q5/Q6 collectors, so to avoid the crossover distortions.
Notice the base of Q1/Q2 are kept fixed, at +/-15V respectively, and this will be the opamp's supply voltage (minus 0.6V drop on each BE junction). Since the Ib of a transistor is much smaller than Ic, the BE current of the Darlingtons Q5/Q6 will see a current proportional with Ic of Q1/Q2, which is the current going through the supply pins of the opamp, which for this configuration is the same as the current going through the zero-ohms virtual load. The Darlington pairs Q5/Q6 will amplify further, and they'll sum their collector currents at the OUT node.
Q3/Q4 are there to dial down the power transistors Q5/Q6 in case of short-circuits or over-current at the circuit's output OUT. D1 and D2 are to protect the opamp against voltage inputs greater than +/-15V.
Finally, let's close the CF loop. From OUT, the Rcf resistor together with Rin form the CF network. Notice the Rin also acts as a terminator for the 50ohms cable, cable that is coming from the signal generator. Notice that Rin is connected in series with the cable, not in parallel.
And that is how it all works as a Current-Feedback Buffered Opamp, and why all the advantages of a CF topology apply to this schematic, too (or at least that is how I assumed it works while designing it, 'cause this is my first current-feedback design

).
The LTspice plots are for a 1MHz square wave input of 1Vpp, and the output produces about 100Vpp into a 50ohms load (+/-1A) parallel with a 1nF.
In comparison, the former voltage-feedback topology was struggling at only 100kHz, while the current CF has no problems at 1MHz, and seems to be going OK up to about 10MHz. I'm very impressed how well it performs in simulation. Will see how well it works in practice, or if works at all.
