The reflection at the end of a transmission line is given by
\$ \Gamma = \dfrac{Z_L - Z_0}{Z_L + Z_0}\$
Where Z0 is the line's characteristic impedance and ZL is the load impedance.
So if you have a 55 Ohm line and you terminate it with 50 Ohms, you're looking at about 5% reflection.
If you have a 50 Ohm line, followed by a short length of 55 Ohm line, terminated with 50 Ohms, you're going to get about 5% reflection from the mismatch between the two lines, and 5% reflection from the end termination.
But...these two reflections will interact with each other. Depending on your operating frequency and the length of the mismatched line, they could interfere with each other constructively or destructively. You could have anywhere from 0 to +/-10% reflection from this structure.
But this is about the voltage signal (or travelling wave) reflection. If you're talking about "losses" you probably want to convert this to power terms. If you know the signal reflection, that would correspond to \$(1-|\Gamma|)^2\$ fraction of the power transmitted to the load. For 10% reflection, that's about 81% power delivery or 0.9 dB of loss.
To find out the exact reflection for your situation, you'd need to either run a simulation, or do a pen-and-paper analysis with a Smith Chart, which we've discussed a little bit in a previous question. The simulator is probably the easier way to go, though, and it will enable you to do things like sensitivity analysis more quickly.
The issue I see if you're trying to create a matched termination, is that except for the one at upper-right, your terminations are all short circuits, not matched terminations.
Since your frequency band is exactly one octave, it's possible that you could design the length of the CPW from your probe pads to the short-circuit to be approximately 1/8 wavelength, so that the short will appear as a match when seen from your probe point. This will work well for a narrow band around, say, 1.414 GHz, and will be a very bad approximation at the edges of your band at 1 and 2 GHz. If you have space, you could make different test structures with different lengths for testing in different portions of your band.
If you can work out how to do it, the option at upper-right would create a matched termination over a much broader band, but as you say it would require very careful design to ensure it's really a broadband 50 Ohm termination. From a geometry p.o.v., I'd suggest using a symmetric structure with 100 Ohm resistance from the central trace to the ground on each side.
An option that might be even better is to build a "through" structure instead of a stub structure. Put probe pads at both ends of your transmission line, and use two probes. Then let the VNA and its 2-port calibration math work out the errors due to the slight mismatch of the probe at the far end, instead of relying on your assumed-perfect 50-Ohm load as a reference for determining the trace impedance.
Best Answer
I was somehow in the same situation recently but my configuration was a differential pair in a stripline mode. I tried different calculators and got different results. After trying to understand how the computation was done by the calculators, I gave up.
I looked for the most recent equations about stripline impedance and made my own calculator in a spreadsheet. Maybe the result is not the most accurate or doesn't take into account some parameters, but as an engineer in a preliminary design, I had to understand the result to be able to see some trade-off. Reading some signal integrity books teach me that the accuracy of these equations (for stripline, but I think it's about the same for CPW), is 10 to 20%.
You can find some equations on CPW impedance here : http://qucs.sourceforge.net/tech/node86.html
To get a better value, a 2D solver is required (accuracy is under 5%). Look at Polar Instruments ( http://www.polarinstruments.com ), their 2D solver is pretty simple but quite accurate. HFSS is a 3D solver, maybe too complex for a first guess of impedance.