You need at least one "current limiting resistor" per LED, and you need to size them appropriately for each LED's forward voltage.
Each LED has a particular "forward voltage". For red, typically 1.8V, while blue/green can be as high as 3.2V. What is happening, is when you add the red LED, it becomes the path of least resistance, due to the (much) lower forward voltage, and all current goes through the red LED.
So, let's say your power source is 4.5V (3x alkaline batteries), and each LED is rated for 20mA of current (pretty typical). If the red LED has a Vf of 1.8V, you need to give it a series resistor of 135 Ohms. If the blue LED has a Vf of 3.2V, then it gets a series resistor of 65 Ohms.
I can get into the math if you'd like (it's not too bad), but there are calculators online too:
These LED+resistor pairs can then be connected in parallel, and your lights will work.
I just read that your power source is a little coin cell battery, these tend to have very high internal resistance, which is why the red LED is not burning out, but rather absorbing all the current. It is acting as though there is a high-value resistor in series with your parallel LED's. When you put diodes in parallel like that (LED stands for light emitting diode), only the one with the lowest forward voltage will conduct, so the red wins here.