Since $e^2$ ends in $e$, $e$ must be $1,5$, or $6$. Clearly $e\ne 1$, since $abcde\cdot 1=abcde$, not $edade$. In fact, $a$ must be $1$ in order for $e$ to be the first digit of the product. Thus, we now have
where $e$ is $5$ or $6$. Next, $be$ does not produce a carry. If $b$ were $2$ or more, $be$ would have to produce a carry, since $e\ge 5$, and $b$ cannot be $1$, so $b=0$, and we have either
In the first case $5\cdot d+2$ ends in $d$; $5\cdot d$ ends in either $0$ or $5$, so $5\cdot d+2$ ends in $2$ or $7$, and you can check that both $d=2$ and $d=7$ satisfy the condition that $5\cdot d+2$ end in $d$.
In the second case $6\cdot d+3$ ends in $d$, and you can check that there is no value of $d$ for which this is true. Thus, $e=5$, and $d$ is $2$ or $7$. The possibilities at this point are therefore
From here you should be able to finish it pretty easily.
Added: I wrote this up exactly as I worked out the problem myself, as an illustration of the thought processes involved and, to be honest, because I didn’t feel like trying to polish it; as a result, it’s not as efficient as it could have been. As Henry notes in the comments, $$10999\cdot5=54995<55000\;;$$
had I noticed this, I could immediately have ruled out the possibility $d=7$.