How does a Wilkinson power divider achieve isolation?
Microwaves101 provides a good qualitative and intuitive explanation, and you seem to understand the idea. For a more rigorous derivation of the scattering matrix you can read section 7.3 (page 328) of Pozar's Microwave Engineering. In this case he uses what is called Even-Odd mode analysis, which he explains pretty well since it is the first time it is used in the text.


  For example, why split in half and not some other ratio which would not result in complete cancellation?


The whole idea of the circuit is that it has the property that ports 2 and 3 are 180 degrees away (through the microstrips) and also 0 degrees away through the resistor. This circuit is the simplest way to achieve this property.


  And wouldn't port 1 have some significance to the operation?


Yes. The circuit is only lossless if all ports are matched. If there are unequal voltages at ports 2 and 3 at any time power will be lost in the resistor. Assuming all ports are matches, this means the circuit can divide a signal losslessly (because the V2 and V3 will be equal and in phase) and can combine identical &amp; in phase signals losslessly. It cannot combine independent signals losslessly*. This makes them a good candidate for use in solid state RF power amplifiers.

*It is impossible to construct a three-port lossless reciprocal network that is matched at all ports in general.