There are three options I know of. One is to terminate one of the ports with 50ohms and take 2-port measurements.
I think it will reduce the math involved if you terminate each port with a load that matches its output impedance. For example if you terminate port 2 with a matched load of Z2, then measure the S-parameters between ports 1 and 3 with a 2-port network analyzer, you will get the correct S13 and S31 because there will be no signal incoming to port 2.
Unfortunately it's not likely you'd be able to make a perfectly matched load for port 2 without something like laser-trimmed resistors.
The second option is to take a 3-port measurement.
If you have the equipment to do a 3-port measurement, that sounds like by far the most straightforward way to solve this.
The problem is, the two output traces are about 10-20 times the width of a 50ohm line, with spacing near the minimum allowed. If I connect to the output ports with a wide line or taper, I'll have moding and coupling errors. If I connect to them with a 50ohm line, I'll get errors due to the large step in width.
If you can, I'd suggest to use lines matched to the device's output ports to connect to the probe pads. Then you will calibrate your network analyzer (VNA) at the probe's reference planes. The VNA will have built in the capability to transform the port impedances, which will take into account the calibrations and make it as if you had measured with a VNA with the desired port impedances.
I'm considering placing two identical splitters back to back and taking a 2-port measurement.
I don't think this will be effective if done the way you show it in your drawing. You'll have problems if you can't perfectly match the phase delay in the lines connecting the port 2's and the port 3's, for example.
Better would be to, for example, terminate port 2 on each of the devices, and connect the port 3's. And measure the 2-port parameters between the port 1's. Then you'll get a measurement of \$S_{31}S_{13}\$ from which you can estimate the actual actual \$S_{31}\$ by assuming the two devices are identical. Again, imperfections on the matching of the port 2 termination will contribute errors.
How is \$S_{21}\$ contained in \$S_{11}\$? By the definition of S-parameters, \$S_{11}\$ is the amount of wave leaving the two-port from port 1 when all other ports are terminated in a matched load, and a wave is incident on port 1. \$S_{21}\$ is the wave energy leaving port 2 due to an incident wave on port 1. Therefore, your original statement is correct, in that the power lost (assuming all energy absorbed in the two port is loss) is:
$$ 1 - |S_{11}|^2 - |S_{21}|^2 $$
The amount of incident wave power, minus the reflected power, minus the power that makes it through the system. In fact, this equation is often used to define a lossless, passive two-port (although it's usually written in a different form, and it has to apply for both directions). One could write this as:
A passive two port is lossless when:
$$ |S_{11}|^2 + |S_{21}|^2 = 1$$
$$ |S_{22}|^2 + |S_{12}|^2 = 1$$
*Usually this is written in matrix equations, where they have the added benefit that they can be generalized to N-port networks.
Best Answer
I'm not sure what you mean about ignoring "the sides". You shouldn't ignore any component in your analysis. Of course, if you find there is no current flowing through a resistor, then the voltage across it will be zero.
The Z parameters allow you to calculate
$$\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}Z_{11} & Z_{12} \\ Z_{21} & Z_{22}\end{bmatrix}\begin{bmatrix}i_1\\i_2\end{bmatrix}$$
meaning if you know the two port currents, you can calculate the port voltages.
So to work out the Z matrix from circuit analysis, you apply a fixed current at each port (one at a time), and see what voltages result.
If you apply 1 A at port 1, and measure a certain voltage \$v_2\$ at port 2, then that immediately gives you the value of \$Z_{21}\$, for example.