= Solution
For fixed $g>0$, write $A=\hbar T/(2m)$. The bound contribution to the <quantum propagator of an attractive delta potential> remains $\kappa e^{i\hbar\kappa^2T/(2m)}$. For its continuum, the <stationary phase> is at $k=0$, where $k^2/(k^2+\kappa^2)=k^2/\kappa^2+O(k^4)$. The Gaussian oscillatory integral and its derivative give
$$
\int_{-\infty}^\infty e^{-iAk^2}dk=\sqrt\pi e^{-i\pi/4}A^{-1/2},\qquad
\int_{-\infty}^\infty k^2e^{-iAk^2}dk=\frac{\sqrt\pi}{2}e^{-3i\pi/4}A^{-3/2}.
$$
Consequently
$$
\boxed{K(0,T|0,0)=\kappa e^{i\hbar\kappa^2T/(2m)}
+\frac{e^{-3i\pi/4}}{4\sqrt\pi\kappa^2}\left(\frac{2m}{\hbar T}\right)^{3/2}
+O(T^{-5/2}).}
$$
The continuum decays as $T^{-3/2}$, but the full kernel has a persistent oscillating <bound state> contribution and no ordinary complex limit. This expansion requires $\hbar\kappa^2T/(2m)\gg1$; it is not uniform in the zero-coupling limit, where the bound state disappears and the free kernel instead decays as $T^{-1/2}$.
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