My understanding is that the introduction of time-ordering in this derivation can be justified if the momenta of all outgoing particles are different from the momenta of the incoming particles. This isn't a serious constraint, since we are dealing with a scattering process, after all!
Having said that, here's how you could justify time-ordering. First, note that equation (5.10) essentially comes from
$ a_1^\dagger (t+dt) - a_1^\dagger(t) = \partial_0 a_1^\dagger(t) \cdot dt$. Formally we could write $$ a_1^\dagger (t_2) - a_1^\dagger (t_1) = \int_{t_1}^{t_2} dt \partial_0 a_1^\dagger(t) $$
Now, we are going to evaluate $ \left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right>$. One can take the term $a_{2'}(t_2)$ and use the above formula to bring it back in time to $t_1$:
$$\left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> = \left<a_{1'}(t_2) \left(a_{2'}(t_1) + \int_{t_1}^{t_2} dt \partial_0 a_{2'}(t)\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> $$
Recall that the commutation relations for operators $a(\vec{k},t)$ is such that the only non-zero equal-time commutator is $[a(\vec{k}, t), a^\dagger(\vec{k'}, t)]=(2\pi)^3 2\omega \delta^3(\vec{k}-\vec{k'})$. Note that for non-equal-time commutators, because the Lagrangian now can have interaction terms, time-evolution will mix both raising and lowering operators with different $\vec{k}$, which is why we brought $a_{2'}(t_2)$ back to $t_1$ to check how it would commute with the term behind it, $a_1^\dagger(t_1)$.
Now, because $a_{2'}$ is centered around $\vec{k}_{2'}$ in momentum space, while $a^\dagger_{1}$ is centered around $\vec{k}_{1}$, we have $\delta^3(\vec{k}_{2'}-\vec{k}_1) = 0$. Therefore all equal-time pairs of $a(t)$ and $a^\dagger(t)$ commute. You now see that $a_{2'}(t_1)$ can move all the way to the right and annihilate the vacuum. The equation now reads
$$\left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> = \left<a_{1'}(t_2) \left(\int_{t_1}^{t_2} dt \partial_0 a_{2'}(t)\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> $$
Next, pick a certain term in the integral, and try to evaluate
$$\frac{\partial}{\partial t} \left<a_{1'}(t_2) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right>$$
You may have guessed that I would bring $a_{1'}(t_2)$ back in time to $t$ to get
$$
\begin{eqnarray}
& & \left<\left(a_{1'}(t) + \int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\
& = & \left<a_{2'}(t) a_{1'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\
& = & \left<a_{2'}(t) \left(a_{1'}(t_1)+\int_{t_1}^t dt' \partial_0 a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \\
& & \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\
& = & \left<a_{2'}(t) \left(\int_{t_1}^t dt' \partial_0 a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \\
& & \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right>
\end{eqnarray}$$
Therefore
$$
\begin{eqnarray}
& & \left<a_{1'}(t_2) a_{2'}(t_2) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\
& = & \int_{t_1}^{t_2} dt \frac{\partial}{\partial t} \left<a_{2'}(t) \left(\int_{t_1}^t dt' \partial_0 a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> + \\
& & \int_{t_1}^{t_2} dt \frac{\partial}{\partial t} \left<\left(\int_{t}^{t_2} dt' \partial_0 a_{1'}(t')\right) a_{2'}(t) a_1^\dagger (t_1) a_2^\dagger (t_1) \right> \\
& = & \int_{t_1}^{t_2} dt \frac{\partial}{\partial t} \int_{t_1}^{t_2} dt' \frac{\partial}{\partial t'}\left<\left(\mathrm{T} a_{2'}(t) a_{1'}(t')\right) a_1^\dagger (t_1) a_2^\dagger (t_1) \right>
\end{eqnarray} $$
After applying the same trick to the $a^\dagger$ terms, we have
$$
\left<a_{1'}(+\infty) a_{2'}(+\infty) a_1^\dagger (-\infty) a_2^\dagger (\infty) \right> =
\int_{-\infty}^{+\infty} dt_{2'} \frac{\partial}{\partial t_{2'}} \int_{-\infty}^{+\infty} dt_{1'} \frac{\partial}{\partial t_{1'}} \int_{-\infty}^{+\infty} dt_{1} \frac{\partial}{\partial t_{1}} \int_{-\infty}^{+\infty} dt_{2} \frac{\partial}{\partial t_{2}} \left<\mathrm{T} a_{2'}(t_{2'}) a_{1'}(t_{1'}) a_1^\dagger (t_1) a_2^\dagger (t_2) \right>
$$
Finally, we invoke the commutability of the time-ordering operator $\mathrm{T}$ and time-derivative $\partial_0$ to get the final form of the LSZ formula.
