Electronic – Why can’t resistors be used to connect cathode pins to GND instead of segment pins to VDD with a 7-segment display

You are not using enough voltage.
The page you referenced notes that there multiple LEDs in series.
They say:

• Large (58mm) 7-segment LED display
  Digits color can be either red or green.
  Type: 12101BEG
  If max: 30mA
  If peak 1/10 dutycycle, 0.1ms: 100mA
  Vf red: 7.2-8.8V
  Vf green: 8.8-10.0V
  Each segment consists of four LEDs connected in series.

In most cases simply trying every combination in turn will work.
There is a small chance that you will destroy segment LEDs with reverse voltage.

Supply two leads from psu via a resistor so you can easily swap both leads.
Try each combination in turn.
A1 A2 A3 A4 A5 B1 B2 .......

The red wire is hooked up to the 5v output on the arduino and the blue to the ground. As can be seen the decimal point has been lit up in green, and if I move the red wire four pins to the left the decimal point turns red instead.

As they say CA = common Anode = common +ve you may have
2 x Anode's
+ 8 x Cathodes.
So Bottom right may be common green anode and
bottom left = common red anode.

SO increase Vsupply to say 8V, use a 10k resistor for initial safety, put V+ on bottom right and try all pins with v-. If that works then put V+ via 10l on bottom left and repeat.

The supplier should be able to provide a datasheet.

Here is a possible mechanism for the fault. You say it works (dimly) without the transistors, but doesn't work with them. So it sounds like the fault is in the transistors. Is there some reason that the transistors might be letting some current through when they shouldn't?

Yes. You're using a PNP transistor. As you know, these transistors are on when the base voltage is lower than the emitter voltage. They are off when the base voltage is higher than or equal to the emitter voltage.

The problem with the shift register chip is that the outputs are always lower than the emitter voltage. I couldn't quite make out the part number of the chip you're using, but according to the datasheet for the 74HC595 (page 6), the outputs don't quite reach Vcc. If there is a tiny voltage difference, then you could find that a small amount of current is leaking out of the base of the PNP transistor. With a gain of about 100, you could find that there's enough CE current to give noticeable light output on the LEDs.

Something to try: Add a schottky diode between Vcc and the emitter. This should drop the collector voltage by a fraction of a volt, just enough to allow the shift register to fully turn off the transistor.