And here is the number you are probably thinking of:
It works only for $ab$ where $a \le b$. I suppose that it is a mistake in the problem statement. Others have proven that as it is, the problem is unsolvable.
And here is how I came to that number.
This behaviour of the multiples being a rotation of the original number occurs when the number is the recurrent part of the decimal expansion of some 1/N.
For example for N=7, 1/N = 0.142857142857... . The recurrent part is 142857. The multiples of 142857 are 142857, 285714, 428571, 571428, 714285, 857142 and 999999.
The additional 00 to add is because 1/N actually starts with 0.009174... The zeroes are stripped in the original number, but must be restored in every rotation of the number. The fact that the recurrent part starts with 00 tells me that N is betwen 100 and 999.
The recurrent part of $1/N$ for a large N is of the form $(10^R-1)/N$ where R is the period of the decimal expansion, or the length of the recurrent part. So I checked only numbers of that form.
The number of digits should have been 100 to accomodate with all the combinations of $ab$ once from 10 to 99, plus a second copy of $aa$'s from 11 to 99, plus 1 because the first and last digits can match only once. The actual length is 106. Since cases with $a<b$ are not counted, it actually could be anywhere from 46 and larger. Anyway I tried all lengths up to 120.
So I searched for numbers of the form $(10^R-1)/N$ that contain all combination of 2 digits $ab$. I didn't find any so I ignored the case $b=0$. I found a few candidates. But checking for which $ab$ the multiplication procedure actually works, I noticed that it works only when $a \le b$. Since other have proven that the problem is not possible as stated, I assume it is a mistake, maybe the procedure for $a>b$ is different than for $a>b$.
PS: I have been playing with this problem. You can extend it to $ab$ with $a > b$ with the following rule:
- if a = b+1 then search $bb9$ and split between the b's.
- if a > b+1 then search $b(a-1)$ and split between b and (a-1).
For example, to multiply by 42 don't search for 24 but 23 and split the number between 2 and 3. For the search with 0's you might need to imagine the '00' in front.