If I understand your question correctly, it's why do we try to fit $\frac{1}{1+e^{-a(x-b)}}$ to our data rather than $\frac{1}{1+\kappa^{-a(x-b)}}$ ? (I've used the univariate case here just to keep things simple)
Consider, that for any (positive) $\kappa$, we can write it as $\kappa = e^{c}$, and thus $\frac{1}{1+\kappa ^{-a(x-b)}}=\frac{1}{1+(e^{c})^{a(x-b)}}=\frac{1}{1+e^{ac(x-b)}}$
So from a solution perspective, it doesn't matter whether you use the natural exponent, or any other, the curve in this function of families which maximises your likelihood will be the same, and your parameter (in the above example, a) will just take a different value, depending on which exponent you use.
This is true for any positive $\kappa$, and for negative $\kappa$, you lose the basic property that your output probability must be between 0 and 1.
So in conclusion, you don't have to use the natural exponent, you can use any positive one, but using the natural exponent keeps gradient calculations simple, and most likely any other follow on calculation you might want to do. You could do it with other exponents, but then you'd have to keep track of trailing logarithms.