First place the vowels in alphabetical order.
$$AIIIOU$$
We now have $3$ $T$'s, $1$ $S$, $1$ $L$ and the $2$ $N$'s (which we shall treat as a group as you suggest).
We will first place the group of $N$'s, the $S$ and the $L$ as they have no restrictions on them.
There are $7$ gaps we can put the $S$ in.
There are $8$ gaps where we can then put the $L$.
Then there are $9$ gaps where we can put the two $N$'s. (sorry for forgetting this last time perhaps confusing you).
Then $10 \cdot 9 \cdot 8$ ways to place the $T$'s. But we have to divide this number by $3!$ as the $T$'s are not distinguishable. So $5 \cdot 3 \cdot 8=\binom{10}{3}\binom{10}{7}$ is the actual number.
So our final result is $7 \cdot 8 \cdot 9 \cdot (5 \cdot 3 \cdot 8)=60480$.
I'm SO sorry for getting it wrong last time. I have a weakness for overlooking things in combinatorics :).
Method 1: We use symmetry.
For the moment, let's consider the number of distinguishable ways we can permute the ten letters in BOOKKEEPER. We can place the three E's in three of the ten locations in $\binom{10}{3}$ ways. We can place the two O's in two of the remaining seven places in $\binom{7}{2}$ ways. We can place the two K's in two of the remaining five places in $\binom{5}{2}$ ways. There are then $3!$ ways of arranging the B, P, and R in the three remaining places. Hence, the number of distinguishable arrangements of BOOKKEEPER is
$$\binom{10}{3}\binom{7}{2}\binom{5}{2} \cdot 3! = \frac{10!}{7!3!} \cdot \frac{7!}{5!2!} \cdot \frac{5!}{3!2!} \cdot 3! = \frac{10!}{3!2!2!}$$
Now, let's restrict our attention to arrangements of the five vowels in BOOKKEEPER. Since there are three E's and two O's, a given permutation of EEEOO is determined by in which three of the five positions the E's are placed. There are $\binom{5}{3} = 10$ ways to do this, of which just one is in alphabetical order.
Hence, the number of permutations of the letters of BOOKKEEPER in which the vowels appear in alphabetical order is
$$\frac{1}{10} \cdot \frac{10!}{3!2!2!} = \frac{9!}{3!2!2!}$$
Method 2: We place the consonants first.
There are $\binom{10}{2}$ ways of choosing the positions of the two K's, eight ways to place the B, seven ways to place the P, and six ways to place the R. Once the consonants have been placed, there is only way to fill the five remaining positions with the vowels in alphabetical order. Hence, the number of distinguishable arrangements of the letters of BOOKKEEPER in which the vowels appear in alphabetical order is $$\binom{10}{2} \cdot 8 \cdot 7 \cdot 6$$
Method 3: We place the vowels first.
There are five vowels in BOOKKEEPER, which has ten letters. We can select positions for the five vowels in $\binom{10}{5}$ ways. There is only one way to arrange the vowels in those positions in alphabetical order. There are $\binom{5}{2}$ ways to place the K's in two of the remaining five positions. There are $3!$ ways to arrange the B, P, and R in the remaining three positions. Hence, the number of distinguishable arrangements of BOOKKEEPER in which the vowels appear in alphabetical order is $$\binom{10}{5}\binom{5}{2} \cdot 3!$$
Best Answer
Your treatment of the Ss is completely right.
You made a minor error in the treatment of the vowels. Yes you can treat them as one type of letter, call it $\nu$, but there are still three instances of $\nu$ in the arangments.
So the problem is equivalent to the number of arrangements of $8$ symbols, two of which (SSS and C) are distinct from all other symbols, and the other six of which consist of two groups of three identical symbols ($\nu\nu\nu$ and TTT).
You then can reason:
$8$ places to put the SSS.
Then $7$ slots to put the C.
Then $\binom{6}{3} = 20$ ways to choose the three slots to put the Ts.
At which point, the slots to put the vowels are fully determined, and there is only one arrangement of the vowels allowed among those slots.
The answer, then, is
$$ 8\cdot 7\cdot 20 = 1120$$