Possible answer:
$ \DeclareMathOperator\sech{sech} $
The standard tractroid is parametrized as
$$(t,\theta) \mapsto (t - \tanh t, \sech t \cdot e^{i\theta}) \in \mathbb{R}\times \mathbb{C} \cong \mathbb{R}^3 $$
One can compute that the unit normal is
$$ \vec{n} = (\sech t, \tanh t \cdot e^{i\theta}) $$
and that the two families of asymptote curves are given by
$$ \gamma_{\pm,\theta_0}(t) = (t - \tanh t, \sech t \cdot e^{i(\theta_0 \pm t)} ) $$
here the parametrization has unit speed. Observe also that for $t$ very negative,
$$ |\gamma_{\pm,\theta_0}(t) - (t + 1,0)| = |(1 + \tanh t,\sech(t))| = 2|1+\tanh t| $$
which decays exponentially as $t\to -\infty$.
In particular, given $t$, choose $\theta_\pm$ such that $e^{i(\theta_\pm \pm t)}$ are equal, the inner product between the two families are
$$ \langle \dot{\gamma}_{+,\theta_+}(t), \dot{\gamma}_{-,\theta_-}(t) \rangle = 1 - 2 \sech^2(t)$$
Note that the corresponding angle is order $2 \sech(t)$ which decays exponentially as $t \to -\infty$.
Thus: suppose you are interested in generating the Amsler surface starting with a pair of line segments with lengths $M$ and opening angle $\theta$. Fixing $M$ and taking $\theta\to 0$ you will find that the sliver of the tractroid bounded by $\gamma_{\pm,\theta_\pm}(t)$, where $t\in [t_0,t_0+M]$ with $t_0$ being the negative solution to $1 - 2\sech^2(t_0) = \cos\theta$, is an approximate solution to your problem with error of order $\theta$.
Your generated picture indicates that you are using fixed grid sizes in your mesh, which are much larger than the small initial angle. If this is the case, probably your numerical solution is not that reliable in the case of the small angles. You probably want to adjust your mesh size to be much smaller than the opening angle, so that you can reliably resolve the difference between the two solutions. (Also, looking at the final picture, I don't see obvious candidates for the two "straight lines" that should appear for the Amsler surface; so perhaps you are not prescribing that as boundary condition but instead solving for it dynamically?)
