We give a detailed argument that is less sophisticated than the argument of Carl Mummert, but that should be sufficient to settle the question.
Let $L$ be the language with constant symbol $0$, unary function symbol $S$, and binary function symbols $+$ and $\times$.
Let $T$ be any consistent theory over $L$ which extends (first-order) Peano Arithmetic and such that every theorem of $T$ is true in the natural numbers, under the standard interpretation of the non-logical symbols. Then the set of sentences of $L$ that are theorems of $T$ is not decidable.
To prove this, we use the result of Matijasevich that every recursively enumerable set is diophantine. Let $A$ be any recursively enumerable set. Then there exist polynomials $P(x,y_1,\dots,y_n)$ and $Q(x,y_1,\dots,y_n)$, with natural number coefficients, such that for any natural number $a$, $a \in A \longleftrightarrow \exists y_1\dots\exists y_n(P(a,y_1,\dots,y_n)=Q(a,y_1,\dots,y_n)).$
Suppose now that $a\in A$. Then there really are natural numbers $b_1,\dots,b_n$ such that $P(a,b_1,\dots,b_n)=Q(a,b_1,\dots,b_n)$. It follows that the sentence $P^\ast(a,b_1,\dots,b_n)=Q^\ast(a,b_1,\dots,b_n)$ is true in the natural numbers. (By $P^\ast$ and $Q^\ast$, we mean $P$ and $Q$ "formalized," with natural numbers replaced by the appropriate numerals, and exponents replaced by repeated multiplications).
Since $P^\ast(a,b_1,\dots,b_n)=Q^\ast(a,b_1,\dots,b_n)$ is true in the natural numbers, it has a very simple proof-by-calculation, that $T$ is certainly plenty strong enough to handle. It follows that $\exists y_1\dots\exists y_n(P^\ast(a,y_1,\dots,y_n)=Q^\ast(a,y_1,\dots,y_n))$ is a theorem of $T$.
Conversely, suppose that $a\notin A$. Then
$\exists y_1\dots\exists y_n(P(a,y_1,\dots,y_n)=Q^\ast(a,y_1,\dots,y_n))$ is false in the natural numbers. By the choice of $T$, this implies that $\exists y_1\dots\exists y_n(P(a,y_1,\dots,y_n)=Q^\ast(a,y_1,\dots,y_n))$ is not a theorem of $T$.
In particular, suppose that $A$ is recursively enumerable but not recursive (there are many such sets). We conclude that the set of natural numbers $a$ such that $\exists y_1\dots\exists y_n(P^\ast(a,y_1,\dots,y_n)=Q^\ast(a,y_1,\dots,y_n))$ is a theorem of $T$ is not recursive. Since the sentences of the above shape are (under a suitable indexing) clearly a recursive subset of the set of all sentences, it follows that the set of sentences provable in $T$ is not recursive.
Comment:
$1$. The result of Matijasevich is not needed to carry out the argument. It has long been well-known that every recursively enumerable predicate $A(x)$ can be expressed in the form $\exists y F(x,y)$, where $F$ is a formula of arithmetic that only involves bounded quantifiers. For any specific $a$ and $b$, whether $F^\ast(a,b)$ is true in the natural numbers can be settled by a finite computation.
$2$. For convenience, we assumed that $T$ is an extension of Peano Arithmetic. However, if $P(a,b_1,\dots,b_n)=Q(a,b_1,\dots,b_n)$ is true, the formal verification that $P^\ast(a,b_1,\dots,b_n)=Q^\ast(b_1,\dots,b_n)$ requires very little axiomatic machinery, and can be carried out in theories far weaker than PA.