Given a language $\mathcal L$ and a structure $\mathcal A$ compatible with it, then we call $\mbox{Th}(\mathcal A) = \{ \varphi \mid \mathcal A \models \varphi \}$ the theory of $\mathcal A$; where $\varphi$ denotes a sentence in $\mathcal L$.
Proposition: If $\mbox{Th}(\mathcal A)$ is recursively enumerable, then it is decidable.
Proof: Suppose we have an algorithm that given $\varphi$ accepts it $\varphi \in \mbox{Th}(\mathcal A)$ and diverges otherwise. Consider the complement of $\mbox{Th}(\mathcal A)$ in the set of all sentences. This set is precisely $\{ \varphi \mid \mathcal A \models \neg \varphi \}$, so we just have to put a negation in front of the sentence, and use our algorithm for the resulting sentence, hence this complement is recursively enumerable. $\square$
So for $\mbox{Th}(\mathcal A)$ the notions of r.e. and recursive are the same (basically because it is complete). Is this right? Is my proof correct? I am learning about logic now, and sometimes I get easily confused, and if the above is true then this seems to be a nice little fact, but I nowhere can find it stated...
Your result is correct, and the proof looks valid.
For coursework, you should probably explicitly call out that you're using the result that if a set and its complement are both r.e. then they are decidable.
(Particularly pedantic teachers may say, "complement with respect to what", so in order to get absolutely all of the details right, you should specify checking that the input is actually a well-formed sentence before you blindly stick a negation in front of it. But most probably you can get away with leaving that kind of busywork for the reader to imagine).