Nice question.
No, you cannot reduce $HALT_{\text{TM}}$ to $E_{\text{TM}}$, although you can reduce $HALT_{TM}$ to $\overline{E_{\text{TM}}}$. Here we are talking about Turing reduction.
$HALT_{\text{TM}}$ can be reduced to $\overline{E_{\text{TM}}}$
Let $\langle M,w\rangle$ be such that $M\text{ is a }TM\text{ and }M\text{ halts on input }w$.
Now let us construct a TM $M'$. On an arbitrary input $u$, $M'$ will check whether $u$ is the same as $w$.
- if it is not, $M'$ will run forever.
- if it is, $M'$ will simulate $M$ on input $w$. Then once $M'$ halts, $u$ is considered as accepted.
We can make the transformation from $M$ to $M$ be algorithmic, thanks to the existence of a universal TM machine.
It is routine to verify that $\langle M,w\rangle$ is in $HALT_{TM}$ if and only if $M'$ is in $\overline{E_{\text{TM}}}$.
$HALT_{\text{TM}}$ cannot be reduced to $E_{\text{TM}}$
For the sake of contradiction, suppose $HALT_{\text{TM}}$ is reduced to $E_{\text{TM}}$ by reduction $f$. Then $\overline{HALT_{\text{TM}}}$ is reduced to $\overline{E_{\text{TM}}}$ by reduction $f$ as well. Since $\overline{E_{\text{TM}}}$ is Turing-recognizable, so is $\overline{HALT_{\text{TM}}}$. Since $HALT_{\text{TM}}$ is Turing-recognizable, we have $HALT_{\text{TM}}$ is decidable, which is known to be false, however. This contradiction shows that $HALT_{\text{TM}}$ cannot be reduced to $E_{\text{TM}}$.