In logic, the completeness theorem in propositional logic consists in showing that a formula is provable if and only if it is a sentential tautology.
Suppose now that I have a set of non-logical axioms on top of tautologies. Does it make sense to look for a weakest form of the completeness theorem in that case?
On
I am not sure what you mean by 'weakest form of the completeness theorem', but here is something to keep in mind:
Yes, completeness for a purely logical system means that the system should be able to derive all logical truths, i.e. all statements that would be considered true in all possible worlds, e.g. $P \lor \neg P$ (in classical logic, at least)
However, if we try to come up with a set of axioms for a particular domain (e.g. some branch of mathematics), then completeness means that for any statement that we can create using the language that we have specified for that domain, you can either prove that sentence or its negation from the axioms.
So, for example, if we want to try and come up with axioms for some basic arithmetic (see e.g. Peano Axioms), we use some non-logical symbols (e.g. the Peano axioms use the symbols $0$, $s$, $+$, and $\times$). With those we can consider all possible sentences that we can create using that language (together with the good old logical symbols of course), e.g. $s(0) + s(0) = s(s(0))$ ('1+1=2'). Given that every such sentence is considered to be either true or false in the domain we have in mind, completeness says that our axioms should be powerful enough to either prove $s(0) + s(0) = s(s(0))$ or $s(0) + s(0) \neq s(s(0))$.
But note that we have no such requirement for logical truths: the statement $P$ is not a logical truth, and neither is $\neg P$.
So, logical completeness is quite different from completeness relative to a specific domain.
At least: this is what I think you are after with your 'non-logical axioms'. If not, I have completely misunderstood your question :(
If I understand the question correctly, your statement of the completness theorem is: If $\models \varphi$, then $\vdash \varphi$.
This is often called weak completeness. In many logical systems, such as the standard systems for propositional logic and first-order logic, we actually have strong completeness: If $T\models \varphi$, then $T\vdash \varphi$.
Here $T$ is a theory, or a set of non-logical axioms. This is usually what people mean when they talk about the completeness theorem.
If $T$ is a finite set, say $T = \{\psi_1,\dots,\psi_n\}$, then strong completeness follows from weak completeness in most logics (as suggested by HallaSurvivor in the comments). The argument goes like this: If $T\models \varphi$, then $\models \left(\bigwedge_{i=1}^n \psi_i\right) \rightarrow \varphi$. By weak completeness, $\vdash \left(\bigwedge_{i=1}^n \psi_i\right) \rightarrow \varphi$. Then $T\vdash \varphi$.
To carry out this proof, all you need is (a) to have the appropriate syntax to form the sentence $\left(\bigwedge_{i=1}^n \psi_i\right) \rightarrow \varphi$, (b) for this syntax to have the appropriate semantics so that the first step of the argument works, and (c) for the proof system to be sufficient for the last step to go through, i.e., to convert a proof showing $\vdash \left(\bigwedge_{i=1}^n \psi_i\right) \rightarrow \varphi$ into a proof showing $T\vdash \varphi$. Again, almost all standard logical systems have these features.
On the other hand, when $T$ is an infinite set of axioms, it's not so clear that weak completeness entails strong completeness. For propositional logic and first-order logic, this is essentially the content of the compactness theorem, which says that if $T\models \varphi$, then there is a finite subset $T'\subseteq T$ such that $T'\models \varphi$. We can then apply the argument above to $T'$.