Step 1: Decide how many times 20 must appear.
For 20 to count as 'the single mode' of five numbers, it has to occur more times than any other value in the list, with no other value tying it. Start by asking how many times 20 can realistically appear.
Step 2: Rule out 20 appearing three or more times.
If 20 appeared three times, the remaining two numbers would sum to $60 - 60 = 0$, forcing both remaining numbers to be $0$. But then the sorted list would be $0,0,20,20,20$, and its median (middle value) would be $20$, not $18$. This contradicts the given median, so 20 cannot appear three or more times.
Step 3: Rule out 20 appearing once or not at all.
The median is $18$, the middle value of the sorted five numbers, and the fourth and fifth values (both $\ge 18$) are the only positions where a $20$ can sit, since the first two values are $\le 18 < 20$. If $20$ appears at most once among the five numbers, then no value in the list can beat a frequency of $1$ unless some other number repeats instead, in which case that other number, not $20$, would be the mode. So a lone $20$, or no $20$ at all, cannot give a valid 'single mode of $20$'.
Step 4: Conclude 20 appears exactly twice.
Combining Steps 2 and 3, 20 must appear exactly twice, occupying the fourth and fifth positions of the sorted list: $d = e = 20$.
Step 5: Solve the remaining two numbers.
The total is $60$, and $c = 18$, $d = e = 20$ account for $58$, leaving $a + b = 2$ for the first two (smallest) numbers, with $a \le b \le 18$.
The integer solutions are $(0,2)$ and $(1,1)$. But $(1,1)$ makes $1$ repeat twice as well, tying with $20$'s count of two, which breaks the single-mode requirement. Only $(0,2)$ survives, since $0$ and $2$ then each appear just once.
Step 6: State the unique valid set.
The only five-number multiset meeting every condition is $\{0, 2, 18, 20, 20\}$, and since we count unordered selections, ignoring permutations, this is exactly one valid way to pick the numbers.
\[
\boxed{1}
\]