I first note that 18 = 1 x 18 = 2 x 9 = 3 x 6. Hence an element of order 18 is either an 18-cycles permutations,or a product of a 2-cycles and 9-cycles or a product of a 3-cycles and 6-cycles .
Since the first two cases are impossible, and the order of a product of a 3-cycles and 6-cycles is 6 (lcm(3,6) = 6), hence I can conclude that there is no element of order 18 in S9.
Now I wonder whether my proof is completed/correct?
The question is whether there are natural numbers with sum $=9$ and least common multiple $=18$. For any prime $p$ and natural $k$, the lcm of some numbers is a multiple of $p^k$ if and only if at least one of the numbers is a multiple of $p^k$. Thus one of our summands must be a multiple of $9$. But then we have in fact a $9$-cycle, which is of order $9$, of course.
Your argument is mostly correct. Here is another way of phrasing your argument.
The LCM of the cycle sizes (note that there may be more than two cycle sizes! you did not consider this) must be $18=2\cdot 3^2$, so there must be at least one cycle that is a multiple of $9$. If it is a $9$-cycle, then there must be no other cycles (since we are in $S_9$), but this has order $9$. Otherwise it is a cycle of length $>9$, which is impossible in $S_9$.
