Let us try and tropicalize $f(x,y)=x+y-1$. Ley $X=V(f)\subset G_m^2$. SO $X$ is $\Bbb{P}^1$ minus $3$ points.
How is this? I understand that $(G_m)^2$ is a torus in two variables. I suppose $V(f)$ refers to the variety corresponding to $f$ (on the torus?). But how is the solution set of $x+y-1$ the whole of $\Bbb{P}^1$ except for three points?
This is from pg 4 of this paper by Melody Chan.
I think the observation is that $V(f)$ is the set of points where the minimum of $(x,y,1)$ is obtained at least twice. Thus with your example you have $V(f)$ to be $$ y\le x=1, x\le y=1, 1\le x=y $$ and it is clear that the region outside of $V(f)$ consists of 3 edges $x=1, y=1,x=y$. I am not entirely sure how this is related to $\mathbb{P}^{1}$ minus three points, unless one consider the three lines as points on $\mathbb{P}^{1}$ or some other definition is used.
I do not think $G_{m}$ is relevant here. In fact I do not think the algebraic group $G_{m}$ is even properly defined in tropical geometry.