= Solution
Take metric signature $(+---)$ and the <Fourier transform> convention $e^{-ipx}$. Introduce an even source $J$ for the gauge field and odd sources $\eta,\bar\eta$ for the <Faddeev-Popov ghost field>:
$$
Z[J,\bar\eta,\eta]=\frac1{Z[0]}\int\mathcal DA\,\mathcal D\bar c\,\mathcal Dc\,\exp\left[iS+i\int d^4x\,(J^{a\mu}A_\mu^a+\bar\eta^ac^a+\bar c^a\eta^a)\right].
$$
At tree level the two-point functions come from the quadratic action. Integrating the gauge kinetic term by parts gives its momentum-space kernel
$$
K_{\mu\nu}^{ab}(p)=\delta^{ab}\left[-p^2g_{\mu\nu}+(1-\zeta^{-1})p_\mu p_\nu\right].
$$
Its inverse follows from transverse and longitudinal projectors: $K=-p^2(P_T+\zeta^{-1}P_L)$, so $K^{-1}=-p^{-2}(P_T+\zeta P_L)$. Completing the bosonic Gaussian square gives $Z_0[J]=\exp[-iJK_F^{-1}J/2]$, with the <Feynman i-epsilon prescription> in the inverse. Two source derivatives and their insertion factors give the <gauge-boson propagator> $D_F=iK_F^{-1}$. Hence
$$
\boxed{\langle\Omega|T A_\mu^a(x)A_\nu^b(y)|\Omega\rangle=\delta^{ab}\int\frac{d^4p}{(2\pi)^4}e^{-ip(x-y)}\frac{-i}{p^2+i0}\left[g_{\mu\nu}-(1-\zeta)\frac{p_\mu p_\nu}{p^2+i0}\right].}
$$
For $\zeta=1$ this reduces to the <Feynman-gauge adjoint propagator>. The longitudinal double pole is understood with the compatible Feynman boundary prescription.
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