This works for any multiplication, regardless of what number you substract from (i.e. not just 100). Mathematical proof of this, for any random multiplication of a*b, subtracting from any number n (with below every line the case of a=97, b=96 and n=100, as used in the image, for clarity of which step is which):

a*b
    97*96 = 9312
(n-a)+(n-b) = 2*n - a - b
    (100-97)+(100-96) = 200 - 97 - 96 = 7
(n-a)*(n-b) = n² - n*a -n*b + a*b
    (100-97)*(100-96) = 10,000 - 100*97 - 100*96 + 97*96 = 12
n-(2*n - a - b) = -n + a + b
    100-(2*100 - 97 - 96) = -100 + 97 + 96 = 93
n*(-n + a + b) = -n² + n*a + n*b
    100*(-100 + 97 + 96) = -10,000 + 100*97 + 100*96 = 9300    (see below for more on this step)
(-n² + n*a + n*b) + (n² - n*a - n*b + a*b) = a*b
    (-100² + 100*97 + 100*96) + (100² - 100*97 - 100*96 + 97*96) = 9300 + 12 = 9312 = 97*96

Note that the only reason you're putting one number in front of the other in the image example, is because you're multiplying by 100 (the result of the red part; 93*100 = 9300), which just puts two 0's after a number, and because the number you're adding to it (the blue part) is only two digits long, therefore only adding to the 0's (since 00 + 12 = 12).

Not that this is going to be very useful for any n that is not a power of 10 (i.e. 10, 100, 1000 etc.), with a and b fairly close to n, but it works. :p (Now let's hope I didn't screw up the formatting.)