Suppose you have photon's arrangement. Since the loads match the line impedance, there is no reflected power on either line, and the voltage on each of the loads will be the same as the voltage at the splitting point (assuming equal propagation times and dispersions, but not equal impedances) for the two lines. And if the voltages are the same, the two points can be connected without any consequence at all. Therefore, the two loads can be replaced with a single load per your formula, and, since there is no reflected power in either of the two lines, the equivalent line impedance must also be according to your formula for this truth to continue to hold.
Another configuration of interest is where the Photon's two loads are connected, but not matched to the lines. What will be the equivalent Zo of the paralleled lines? To see this, just imagine what you have when the two lines are infinitely long: an impedance that is Z1 || Z2, not dependent on the loads, which matches your formula.
Such arrangements have been used by radio hams since the advent of coaxial cable for impedance matching when cable of the desired impedance (25 or 37.5 Ohms) did not exist.