= Solution
Eliminate $a$ using the <Poynting–Robertson drag invariant>, $a=Ce^{4/5}/(1-e^2)$. The eccentricity rate becomes
$$
\dot e=-\frac{5A}{2C^2}e^{-3/5}(1-e^2)^{3/2}.
$$
Since the point-star limit corresponds to $a\to0$ and $e\to0$, the <inspiral time under Poynting–Robertson drag> is
$$
\boxed{t(e_0)=\frac{2C^2}{5A}\int_0^{e_0}\frac{u^{3/5}}{(1-u^2)^{3/2}}\,du,\qquad C=a_0(1-e_0^2)e_0^{-4/5}.}
$$
For $e_0\to0$, the integral is $(5/8)e_0^{8/5}+o(e_0^{8/5})$, recovering $a_0^2/(4A)$ despite the apparent singularity of $C$.
For $e_0\to1$, use $C\simeq2q_0$ and $1-e_0=2q_0/(Q_0+q_0)\simeq2q_0/Q_0$. The endpoint asymptotic integral is $[2(1-e_0)]^{-1/2}$, giving
$$
\boxed{t\simeq\frac4{5A}\,Q_0^{1/2}q_0^{3/2}.}
$$
Most of this long lifetime is spent near the initially large <apocentre distance>, before substantial circularization. The result assumes weak drag and a valid orbital average; near-parabolic release can violate that assumption.
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