How to Tropicalize a Polynomial in Two Variables? Trying to draw the Amoeba
With Mathematica, it's possible to graph $e^{-k x} + e^{-k y} = 1,e^{-k x} - e^{-k y} = 1$ and $e^{-k x} + e^{-k y} = -1$ to get the amoeba of 1 + x + y when k = 1. Then by plotting when $k \to \infty$, these graphs converge to the tropical polynomial 
http://www.freeimagehosting.net/uploads/7a0cc0b6f3.jpg
However, I meant to draw Min(1,x,y) which I fixed by shifting my coordinates x,y by the vector (1,1).  This is a bit ad-hoc and I am probably not understanding how constant functions "tropicalize".
Main question
I would like to "fatten" Min[x, y, 1, x + y + 1] into its amoeba, so I thought the right curve should be 1 + x + y + xy.  However, my "neck" is disappearing in the scaling limit.  How should I scale the coefficients correctly to get my amoeba?
Following Zeb's suggestion (but a few second before he posted it) I came up with this imge
http://www.freeimagehosting.net/uploads/b612de3bf9.gif
However, this "dequantization" procedure doesn't always produce the whole tropical curve.  Here's the curve I drew to "requantize" Min[1, x , y , x+ y + 1, -2 + 2x ,  2y].  A line has to be missing b/c of the zero tension condition, as in Tropical Mathematics by David Speyer and Bernd Sturmfels.
http://www.freeimagehosting.net/uploads/50219e2142.gif
Here is the code.  You have to draw 4 different versions of the curve to get all the absolute values.  Maybe this should somehow involve complex phases as well.
q[x_] := E^( x)
f[a_, b_, c_] := c + a + b + c a b + (1/c^2) a^2 + b^2;
{x0, y0} = { -1, -3};
L = 5;
k = 8;
ContourPlot[
    { f[q[k x], q[k y], q[k ]] == 0, f[-q[k x], q[k y], q[k ]] == 0, 
    f[q[k x], -q[k y], q[k ]] == 0, f[-q[k x], -q[k y], q[k ]] == 0}, 
    {x, x0, x0 + L}, {y, y0, y0 + L}
]

Ideally, I want to take any tropical curve and fatten it into its amoeba.  Tropical conics and cubics seem the best starting point.

In anticipation of comments, by "amoeba" here I think I mean the boundary of the 2D region which is usually called "amoeba".
 A: For the first amoeba you mentioned, I think your equations should be $e^{-kx}\pm e^{−ky}=\pm e^{-k}$, not $e^{-kx}\pm e^{−ky}=\pm e^{0}$.
For the main question, I think you should be using an equation like $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} = 0$... so really you want a curve like $c+x+y+cxy$, where when you rescale $x$ and $y$ by raising them both to a power, you also rescale the coefficient $c$ by raising it to the same power.
Edit: Ok, for the second problem you are having, I think this is coming up because plugging in different signs of $x$ and $y$ doesn't give you all the different sign possibilities for your detropicalized polynomial.
So, if you want to get the tropical curve $Min(1,x,y,1+x+y,2x-2,2y)$, you want to use all of the equations $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} \pm e^{-k(2x-2)} \pm e^{-k(2y)} = 0$.
In fact, I think you don't need to use all $32$ of those equations, you just need to use enough of them that every pair of terms have opposite signs in one of your equations, such as the following three:
$e^{-k}+ e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} - e^{-k(2y)} = 0$
$e^{-k}- e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} + e^{-k(2x-2)} - e^{-k(2y)} = 0$
$e^{-k}+ e^{-kx}- e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} + e^{-k(2y)} = 0$
In the limit this will give you the right amoeba, but I'm cheating a bit, because really we should be doing fancy stuff with logarithms of complex numbers.
