Technically that should work, you will get very uneven current distribution without a feedback mechanism, but it will light the LEDs.
The problem with what you are suggesting is that for the blue LED you are working with the steep part of the I-V curve of the diode. What that means is a tiny change in the voltage will cause a huge change in the current through the diode.
For example, your batteries will likely start out around 3.2V total for a pair of fully charged alkaline batteries. With a 1 ohm resistor as you suggest, and a 3V diode, the current will be (3.2V - 3V)/1ohm = 200mA. That will burn out a low power LED. Once the supply voltage drops to 3V, the current limiting will be dominated by the LEDs I-V curve and depending on the diode will probably be around 20mA. You can see in this example that a drop in the supply voltage of 0.2V will cause the current to change by a factor of 10.
If you have a steady supply voltage you can size the resistor correctly but even a tiny drop in the charge of the battery can have a large impact on the LED brightness, so not a very good design for anything but demonstration purposes.
What is the purpose of the circuit? Will it need to operate for extended periods of time?