Only reason I ask / want to do this is i have a lot of leds that need
power but don't have enough of the right resistors.
There may be a simple solution. As I understand, you have a 5V supply and need 200 Ω per LED. That means you will drive the LEDs with 15 mA. The problem is you have 100 Ω resistors so have to double the number of resistors.
If the above is true, then all you need to do is add a 3.5V LDO and use 100 Ω resistors. The all LEDs will get 15 mA using a single 100 Ω resistor supplied with 3.5V rather than 5V.
What I am not sure about in your circuit is whether D2 and D3 will ever or never both be on at the same time. So I do not know your desired current.
The above 15 mA may not be exactly what you want but you can drop the 5V to the required voltage to get the desired current with a single 100 Ω resistor for each LED.
I know that Leds in parallel with one resistor is bad.
This is not really applicable to your situation. When the LEDs are in parallel it does not matter much when each resistor is the same value. The concern with parallel LEDs applies more to parallel strings of multiple LEDs. It's the mismatch in the LEDs forward voltage that causes an imbalance. With single LEDs in parallel the imbalance will not very likely be significant enough to be perceivable to the human eye.
If you truly wanted each LED to get the precise same current you would have to choose each resistor value based on the forward voltage of each individual LED. This is a moot point becasue the human eye is not very good at distinguishing differences in luminous intensity between two sources of light.
I have experimented with strips of LEDs in parallel driven with a 2 A constant current driver. One strip draws 800 mA and the other 1200 mA even though the forward voltage were very close e.g. 42.2V and 41.95 when powered individually with 1 A.
The point being 800 vs 1200 is a significant imbalance but it was not perceptible and had to be measured.
In your case the single resistor, R1, should be eliminated if you must use a 5V supply. If you have two 7 segment digits side by side one displaying a one and the other displaying an eight, there may be a perceptible difference in luminous intensity between the two digits if R1 remained in place. This could be true if your R1 was limiting the current for all seven segments.