Say we have a Wilkinson power divider:
So far as I understand it, if we say port 1 is the input, and the other ports are outputs, then the power is split evenly between the outputs.
If one of the outputs is not perfectly matched, and so reflects some power back, half this power is dissipated in the resistor, half goes out port 1, and none (ideally) of it goes to the other output port.
However, does this port isolation hold true if port 1 isn't driven by a Z0 source? And if not, why and how?
No, unfortunately the isolation is not maintained.
The reason you get very little signal at Port 3 from a Port 2 signal is that normally the amount of signal that goes through the 100 ohm resistor is about the same as the net amount of signal which comes back through both transmission lines. About 2/3 of the signal goes through the the first line and about half of that (1/3 of the original) comes back through the other line. These (approximately!) 1/3s are 180 degrees out of phase (two 90 degree lines) and cancel. Of course there is a reflection at each junction and this only works out in steady state.
In any case, if the impedance at port 1 is not 50 ohms, more or less that 1/3 of the original signal will be transferred down line 2. This will either not entirely cancel or over-cancel leading to a net signal out of port 3 which is based on the signal at port 2.
If port 1 is not driven by a source of impedance \$ Z_0 \$ then there will be reflected power from port 1, however the isolation between ports two and 3 will still theoretically be infinite at the frequency where the two transmission line sections of impedance \$ \sqrt{2} Z_0 \$ are \$ \lambda/4\$. The voltage reflection coefficient for the driving impedance \$Z_\alpha\$ will be: \begin{equation} \Gamma = \frac{Z_0 - Z_\alpha}{Z_0 + Z_\alpha} \end{equation}
