Recently, I stumbled across this 10 years old Phys.SE post answered by @Qmechanic him/her self. The question was basically this:
"what happens if we apply time-ordering to Wick's theorem for two operators $$T\{ab\} = N\{ab\} + \langle 0\vert T\{ab\} \vert 0\rangle \ ,$$ wouldn't we get $$T\{ab\} = T\{ab\} + \langle0\vert T\{ab\}\vert 0\rangle \ ,$$ which is a contradiction?".
The answer given is basically that the inner ordering wins on the outer ordering, instead of the opposite being true, thus applying time-ordering to Wick's theorem does nothing.
I faced this same problem a while ago and gave myself a completely different answer. I will therefore explain what I think, hoping that you can either confirm or disprove my point of view. I think the correct explanation of why the contradiction doesn't exist is that you cannot apply time ordering to operator-identities. Time ordering is a notation not a function from operators to operators. Wick's theorem is an operator identity thus the contradiction cannot be reached because applying time ordering to it is a not well defined procedure.
Lengthy discussion
What you could do instead is applying Wick's theorem to a time ordered sequence, which is different. $$ \begin{gathered} T\{ab\} = N\{ab\} + \langle 0\vert T\{ab\}\vert 0\rangle \\ T\{T\{ab\}\} = N\{T\{ab\}\} + \langle 0\vert T\{T\{ab\}\}\vert 0\rangle \\ T\{ab\} = N\{ab\} + \langle 0\vert T\{ab\}\vert 0\rangle \end{gathered} $$ In this last step I used the fact the the outer ordering wins on the inner ordering, in contradiction with what's stated in @Qmechanic's answer. At the end I give a proof of what I say and discuss why $T\{N\{ab\}\} = N\{ab\}$ is not applicable.
The fact that applying Wick's theorem to a time-ordered sequence gives back Wick's theorem again, is not a surprise at all if you consider why Wick's theorem for a product of two operators is true. We know that commutators/anticommutators of creation and annihilation operators are just numbers (operators proportional to the identity) and that a normal ordered sequence always has zero expected value on vacuum, thus it's trivial that $N\{ab\} = ab - \langle0\vert ab \vert0\rangle$. Wick's theorem for $n=2$ is reached by applying the equality to a time ordered sequence (not by applying time ordering to the equality). $$ \begin{gathered} N\{ab\} = ab - \langle0\vert ab \vert0\rangle \\ N\{T\{ab\}\} = T\{ab\} - \langle0\vert T\{ab\} \vert0\rangle \\ N\{ab\} = T\{ab\} - \langle0\vert T\{ab\} \vert0\rangle \\ \end{gathered} $$
Proof that the outer ordering wins. We want to prove that $$N\{T\{A_1 A_2 \cdots\}\} = N\{A_1 A_2\cdots\}.$$ We start by applying the definition of time ordering. $$ \begin{gathered} T\{A_1 A_2 \cdots\} := (-1)^{\text{#fermionic swaps in }P}\cdot A_{P_1} A_{P_2} \cdots \\ N\{T\{A_1 A_2 \cdots\}\} = ((-1)^{\text{#fermionic swaps in }P})^2 N\{A_1 A_2 \cdots\} = N\{A_1 A_2\cdots\} \end{gathered} $$ $P$ is the permutation required to reach time ordering. We used the rules of normal ordering to play in reverse the permutation $P$. A similar proof holds to show that $T\{N\{\cdots\}\} = T\{\cdots\}$.
In the answer given, the opposite fact $T\{N\{ab\}\} = N\{ab\}$ is proven using Wick's theorem for the product of normal ordered sequences. The problem is that the theorem holds true only if all operators that belong to the same normal ordered sequence are calculated at the same time (like always happens in perturbation theory). If this wasn't the case the time ordering of the normal orderings wouldn't even be defined, as we wouldn't be able to associate a single time to each term of the product we are trying to time order. Under this requirements it's trivial that $T\{N\{ab\}\} = N\{ab\}$ since time ordering one term does nothing. Note that this is not in contradiction with $N\{T\{ab\}\} = N\{ab\}$ because $T\{ab\}$ is not defined if $a$ and $b$ are calculated at the same time.
