Which way round should D3 go?
Well, you are right, in DigiSpark's schematic USB power pin becomes useless.
If you look into schematics of some common development boards which run into the same problem of USB supply competing with possible external supply, you'll find either this



(taken from https://www.arduino.cc/en/uploads/Main/Arduino_Nano-Rev3.2-SCH.pdf)

or this



(taken from https://www.arduino.cc/en/uploads/Main/arduino-uno-schematic.pdf)

The Nano's schematic speaks for itself here: the diode's purpose is being a supply-selector. Simply judging based on this fact, we can deduce that it should be connected so that it allows current to flow when there is no external supply, and blocks it if there is one. This implies connection of the cathode to 5V.

Small form-factor boards tend to use a single diode to save space and put up with its disadvantages like relatively large voltage drop and inability to switch the board to 3.3V operation without sinking current from USB. However, even if we look at the body diode of the MOSFET in Uno's schematic, we'll see that it's oriented the exact same way we've just deduced.

Moreover, this arrangement also protects USB bus from possible overvoltage (due to possible connection mistakes). In DigiSpark's schematic the diode does not protect USB from anything at all, it won't save the bus even from, let's say, 7V, because the diode voltage drop is approximately 0.7V (depends on forward current).

I don't see any point in protection of the 7805 from VBUS at all. Under what circumstances could VBUS become >5V +/- 5%? If you really have to take protection to this level, you should look into something like https://www.rugged-circuits.com/ (though I don't quite remember if they offer real USB isolation)

UPD: I can't think of a way to supply current (at least more than a few microamps, if we are talking about a small signal/power diode) through a reverse-biased diode. The fact that it does power from USB suggests that either there is some circuitry around the diode, that lets the current through, or this is actually not a diode.
Are there any board layout files available? Otherwise you have to verify the schematic (just trace the circuit with a multimeter), or try to find datasheet based on the "diodes" SMD marking.
Also you can try to measure current consumption of the device through USB connector, though I think it's highly unlikely that the uC is supplied by reverse current of the diode. But I am not saying it's entirely impossible.

UPD2: I thought I sorted it out, but no. I've come across a photo on some forum links you included that shows that only VBUS pin is required for powering. Data lines are not connected. But anyway, here is what I thought.
During idle state USB D+ line is pulled high. Usually that's what the pullup resistor of the slave device is for, but since facts suggest that the uC is powered even with VBUS blocked, the hub must be doing the same thing. Powering uCs through IO's CMOS protection diodes is a common thing.

UPD3:
As far as I can see the diode is marked "K12". Do the dimensions correspond to SOD-123? If so, then it's a (of course, they have low forward voltage!) Schottky diode, and Schottky diodes are famous for large reverse currents. This [ https://img.ozdisan.com/ETicaret_Dosya/468036_1760701.pdf ] datasheet claims 0.1-3mA, depending on conditions. This may be very well sufficient to power the uC, but can lead to problems when uC's pins are loaded. You can try loading a pin (set to output and high) with a 125-150 Ohm resistor and measuring +5V voltage. If uC is supplied by reverse current, then the latter should drop (and may be even reset the uC doing so).