RF 50 ohm dummy load PCB design
The biggest problem you have is the parallel connection of all those 560 &Omega;  resistors. You obviously need to space them apart so that they can all dissipate their heat. However to parallel them at RF, you need 560 &Omega; transmission lines, and you can't get impedance that high on microstrip, especially the very thin dielectric you've chosen. The eleven lines in parallel will cause a very large shunt capacitive defect, giving you an increasingly degraded return loss as the frequency rises.

This is how I'd approach the circuit







simulate this circuit &ndash; Schematic created using CircuitLab

Resistors are chosen from the E24 series, nearest to the value that would actually give 50 &Omega; match and equal power dissipation. As a result of that approximation, the final DC resistance comes out to about 49 &Omega;. Each horizontal line is a length of transmission line, with the approximate impedance (rounded to integer) above it. You'll note they are all 'reasonable' values, in the ballpark of 50 &Omega;, so straightforward to fabricate on the board. As transmission lines, they can be any length, allowing you to space the resistors at will around the board.

There are some practical details that will affect the high frequency match. High value resistors tend to be a bit capacitive, low value resistors a bit inductive, so this layout would only be a first cut, and measurement, or detailed modelling with data for those specific devices, would be needed to tweak up the match. There is plenty of scope to change the impedance of the series lines slightly to compensate for the parasitics of the components.

Note that the signal decreases along the chain, so R11 provides the ideal place to feed the final instrumentation attenuator, with a slight modification of value as required.