$\begingroup$

You're asking for a theory $T$ such that (1) $T$ is recursively axiomatized, (2) $T$ extends $PA$, (3) if a Turing machine $M$ halts, then $T$ proves that $M$ halts, and (4) for every Turing machine $M$, either $T$ proves that $M$ halts, or $T$ proves that it doesn't. We will argue that no such $T$ exists by way of the well-known

$\bf Theorem\ of\ Kleene:$ There is a pair of computably enumerable sets $A, B$ of natural numbers such that $A \cap B = \emptyset$, and such that for no computable set $C$ do we have $A \subseteq C \subseteq \bar B$, where $\bar B$ is the complement of $B$. Such a pair $A,B$ is called $\it computably\ inseparable$.

Fix $A$ and $B$ as in the theorem. We claim that if your $T$ exists, then we can exhibit a computable $C$ of the kind forbidden by the theorem, a contradiction. For each $n$, let $M_n$ be a Turing machine which halts if $n$ enters $A$ without first entering $B$, and let $M_n'$ be one that halts if $n$ enters $B$ without first entering $A$. (Although $A$ and $B$ are disjoint, the 'without first' clause can still come into play in nonstandard models.) Now fix $n$; we will decide whether to put $n$ into $C$ or into its complement $\bar C$. Since $T$ is consistent, it doesn't prove that both $M_n$ and $M_n'$ halt. If $T$ proves $M_n$ halts, put $n$ into $C$; if $T$ proves $M_n'$ halts, put $n$ into $\bar C$; and if neither of these happens, then by design of $T$, it proves that both fail to halt, and we arbitrarily put $n$ into $C$.

Doing this for every $n \in \mathbb{N}$, we arrive at a computable $C$. We claim that $A \subseteq C$ and $B \subseteq \bar C$. To see $A \subseteq C$, fix any $n \in A$, and notice that $M_n$ halts, so that by choice of $T$, $T$ proves that $M_n$ halts, and therefore we placed $n$ into $C$ in the construction. The proof that $B \subseteq \bar C$ is similar. Thus $C$ is a computable separator for $A$ and $B$, as forbidden by the theorem above, and we are done.