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 indicate that 1/N actually starts with two zeroes: 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, using a computer, 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.