As the question suggests, it is not too hard to show a reduction from the halting problem (HP) to the language $A$ stated in the question, $HP \le A$.
This means that $A$ is undecidable, $A\notin R$ (also, it is not in co$RE$). But, is it in $RE$ or not?
Lets show that $\overline{HP} \le A$ and conclude that $A\notin RE$.
The reduction goes like that: on input $(\langle M\rangle,x)$ we generate the TM $M_x$ that, on input $w$:
- If $w$=$\varepsilon$, ACCEPT.
- Run $M$ on $x$. If M halts, ACCEPT (otherwise, we're in a loop anyways)
The reduction is valid: if $(\langle M\rangle,x) \in \overline{HP}$, then $M$ does not halt on $x$. This means that $M_x$ accepts only the empty string (and loops on any other input). Thus $\langle M_x\rangle \in A$.
On the other hand, if $(\langle M\rangle,x) \notin \overline{HP}$, then $M$ halts on $x$, and $M_x$ accepts any input, thus $\langle M_x\rangle \notin A$.
It is easy to verify that the reduction is computable and complete.