Sure. You can also replace an antenna with a 50 ohm resistor, it looks perfect on the VNA. And a coax cable can also be replaced with just a resistor on one end and then regular cable. I dont even know why people spend so much time designing these things, it is so simple.
Be aware that this is the beginner's section. Your sarcasm may lead the OP to the wrong direction.
To answer the original question: no a 50R trace is not a resistor, but a transmission line with a characteristic impedance of 50R. Look up transmission lines and characteristic impedance.
Even if you build a proper transmission line, it needs to be terminated with 50Ω to ground at the receiving end anyway, unless the load has built-in termination or it's some external RF connector.
Supposedly you may get away with wrong line impedance if the distance is some 10x shorter than the wavelength. But wavelength at 1GHz is merely 20~30cm so that doesn't buy you much. Even is such case, make it a proper transmission line, with a nice ground return conductor running in parallel at uniform distance.
50 ohm traces are going to be extremely wide on a 2-layer 1.6mm board, but you can simply bump it up to a 4-layer PCB. The latter are not that expensive anymore.
I don't think you need to use special substrates like Rogers, unless your application requires really tightly controlled impedances, which I doubt, considering the question. Regular FR4 is going to be just fine up to a gig.
Yes I have read about the 4 layer sollution before but did not understand it at all, could you please explain how 4 layer helps and please on how to layer it? :-)
I do wish people would stop saying how cheap 4 layer boards are... my GPS PCB cost about £3 in two layer. It would have cost more like £43 in 4 layer from JLC, by the time I'm stung for tax and fees on top of the already 10x higher cost.
Have you considered using a semi-rigid subminiature 50 ohm coax instead of a line built into the board? it would make the board simpler and cheaper. RG405 is only .086' outside diameter and there is also some that is .047" in diameter. Cut to length, form, and solder the ends of the shield to the ground plane on the board. Neat and simple.
0.8mm thickness is indeed flimsier. Whether it's a problem depends on how it will be treated. The problem is that much flexing is likely to crack solder joints of course, and possibly crack MLCC capacitors which tends to have very unfortunate results.
If the PCBs are fixed into the enclosure such that flexing cannot occur, as mine are, then there's no problem. If that can't be done then 0.8mm is a poor idea and you'll need 1.6mm. In that case, if you can't make the traces wide enough on 2-layer to get down to 50Ω, then the above solution of using some thin coax is a workable one.
Well it's ArthurDent's suggestion, but I agree with it. Yes, your diagram has the gist of it.
Whether it's better for your design than a 2.88mm wide trace (which is what you need to get 50Ω on a 1.6mm thick 2-layer board)... only you can tell. I had too much going on close by to put such a wide trace in.
I might end up using the coax method on my next PCB because, other than the impedance requirement on multiple tracks, I have no need to go to four layers. It's too big a board and will have a lot of PCB-mount sockets to risk 0.8mm thickness with. I still need to think about the best way to deal with it.