This is partly a slight variant of Henry’s answer and partly a further explanation in answer to one of your questions.
If $d$ is one of the allowable digits $1,3,5,7,9$, let $d' = 10 - d$; note that $d'$ is also allowable. Now let $abcde$ be any allowable number; then $a'b'c'd'e'$ is also allowable, and it must be a different number from $abcde$. (It would be the same number only if all of the digits were $5$, but that’s not allowed.) In this way you can pair up all of the allowable numbers. Since there are $5! = 120$ permutations of the $5$ allowable digits, there are $120$ allowable numbers, and hence there are $60$ of these pairs.
It shouldn’t be too hard to see that $abcde + a'b'c'd'e' = 111110$ no matter which allowable number $abcde$ you start with. That is, the numbers in each pair sum to $111110$. Since there are $60$ pairs, the grand total is $60 \cdot 111110 = 6,666,600$.
Henry did essentially the same thing, except that instead of looking at the sum of $abcde$ and $a'b'c'd'e'$, he looked at their average, $111110/2 = 55555$. Since all pairs have the same average, the entire collection of allowable numbers must also have that average. And if the $120$ allowable numbers have an average of $55555$, their total must be $120 \cdot 55555 = 6,666,600$.