jmc

Starting with the the rightmost column would be the easiest (we'll call it the first, the second rightmost column the second, and so on). Let's consider the possible values for $E$ first.

Since the values must be non-zero, we'll start with $1$. If $E$ is $1$, then $S$ would be $2$ and nothing would carry over. However, since $H+H$ must equal $E$ if nothing carries over and $H$ must be an integer, $E$ can not equal $1$.

If $E$ equals $2$, then $S$ must equal $4$. $H$ would then equal $1$. This satisfies our original equation as shown: \begin{array}{c@{}c@{}c@{}c} &4&1&2_5\\ &+&1&2_5\\ \cline{2-4} &4&2&4_5\\ \end{array}Thus, $S+H+E=4+1+2=7$. In base $5$, the sum is then $\boxed{12_5}$.

Note: The other possible values for $E$ can also be checked. If $E$ equaled $3$, the residue $S$ would then equal $1$ and $1$ would be carried over. Since nothing carries over into the third column, $1+H+H$ would then have to equal $3$. However, $H$ can not equal $1$ because the digits must be distinct.

If $E$ equaled $4$, the residue $S$ would then equal $3$ and $1$ would be carried over. $1+H+H$ would then have to equal $4$, but $H$ can not be a decimal. Therefore, the above solution is the only possible one.