The problem is that you won't be able to turn on the NFET in your configuration. If your positive supply is 5V and your gate is 5V, then the NFET will always be in saturation. You need a large gate-source voltage for FETS to work like switches. Pulling the input of the FETs to 0V probably lights one of the LEDs, because the gate-source voltage of the PFET is 5V.
If these are two separate LEDs, then the easiest way to solve this is to separate the two circuits, and make one LED driven with the PFET, and the other LED driven with the NFET (see schematic below). This allows the best efficiency with the least number of parts.
This solution will work if you have separate LEDs:
If it is an anti-parallel diode, there are still a few options. If the device driving the pin can source and sink the current needed, then you could use the two FETs to act as the opposing side of an H-bridge, as shown below.
If you have an extra pin that can drive the current needed, you could just hook the LED with resistor across two pins, and this would probably give you the simplest board design.
If you're not hooked on the idea of using discrete FETs, then you could consider a dual inverter chip, which would be cheap and easy to use, such as this part.
If you need to drive both LEDs to the same current (e.g. 20mA) then you will probably need a H-bridge to give you the flexibility of adding separate load resistors for each diode.
A rule of thumb for MOSFETs is that when you want them to work like switches, you need to connect the source to ground (NFET) or VDD (PFET). Otherwise, the transistor will start to turn itself off as the source voltage rises.
First off, the precision of your resistance value is going to be limited to the resistor's rating. Consider the list of standard values for resistors with a 5% tolerance. Any number between 10 and 91 is going to be within 5% of one or two of the preferred, E24 numbers. Your perfect resistor might even be in your supplies, labeled as one of these.
E24 ( 5%): 10 12 15 18 22 27 33 39 47 56 68 82
11 13 16 20 24 30 36 43 51 62 75 91
Trying to combine resistors in series or parallel will also combine their possible errors, so you may not be getting much closer. Make sure to measure the resistance of any such construction at the end.
See also:
https://en.wikipedia.org/wiki/Preferred_number#E_series
Best Answer
If you don't need to control the LEDs individually, you can connect the LEDs in parallel, but you need a current-limiting resistor in series with each LED, like so:
simulate this circuit – Schematic created using CircuitLab
The resistor value can be calculated using Ohm's Law. You will need to decide on a suitable current for the LED (NOT the Absolute Maximum value from the datasheet! - 10 mA is safe for common LEDs). Determine the forward voltage drop of the LED, and subtrract that from the supply voltage to get the voltage across the resistor.