OP's question is about the long-time behavior of imaginary-time correlation functions in general. But in fact, the correlation function is ill-defined if the time difference $|\tau_A-\tau_B|$ is larger than the inverse temperature $\beta$.
To see this, suppose that the Hamiltonian $H$ has eigenstates $\{|n\rangle\}$ and associated eigenvalues $\{E_{n}\}$, and assume $\tau_A > \tau_B$ . Then, \begin{equation} \begin{split} &\mathrm{Tr}\big\{\mathscr{T}[e^{-\beta H}e^{\tau_{A}H}A(0)e^{-\tau_{A}H}e^{\tau_{B}H}B(0)e^{-\tau_{B}H}]\big\}\\ &=\sum_{n,n^\prime}\langle n|e^{-\beta H}e^{\tau_{A}H}A(0)|n^\prime\rangle\langle n^\prime| e^{-\tau_{A}H}e^{\tau_{B}H}B(0)e^{-\tau_{B}H}| n\rangle\\ &=\sum_{n,n^\prime}e^{-(\beta-\tau_A+\tau_B)E_n} e^{-(\tau_A-\tau_B)E_{n^\prime}}\langle n|A(0)|n^\prime\rangle\langle n^\prime|B(0)|n\rangle. \end{split} \end{equation}
The energy eigenvalues $\{E_{n}\}$ of a physical system are bounded below and unbounded above. Hence, for both $e^{-(\beta-\tau_A+\tau_B)E_n}$ and $e^{-(\tau_A-\tau_B)E_{n^\prime}}$ to well behave, we must have $0\le\tau_A - \tau_B\le\beta$. Similarly, if $\tau_A < \tau_B$, we have $0\le\tau_B - \tau_A\le\beta$.
Combining the two cases $\tau_A > \tau_B$ and $\tau_A < \tau_B$, it follows that the imaginary-time correlation function is well-defined only if \begin{equation} |\tau_A - \tau_B| <\beta. \end{equation}