I was thinking something like this.
Some kind of number equipped with multiplication which behaves like returning 1 if the largest prime factor (with exponent $\geq 1$) has exponent divisible by two
For example $$\left[\begin{array}{c}2\\3\\5\end{array}\right]\left[\begin{array}{ccc}2&3&5\end{array}\right] = \left[\begin{array}{ccc}4&6&10\\6&9&15\\10&15&25\end{array}\right]$$
And then calculating largest prime exponents:
$$ans = \left[\begin{array}{ccc}2&1&1\\1&2&1\\1&1&2\end{array}\right]$$
And divisibility by 2:
$$ans = \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$$
I doubt it would be a group. How would it behave algebraically?
If "anything" is assocative it's impossible.
Consider two first elements of $b$ then we have $b_1^2=1, b_2b_1=0, \ldots\Rightarrow 0=b_2b_1\cdot b_1=b_2b_1^2=b_2\cdot1=b_2\Rightarrow bb^\top\ne I.$
If "anything" is non associative (multiplication of matricies can be well defined although it will be nonassociative) it's possible.
Let $0, 1, a, b, a+b, 1+a, 1+b, 1+a+b$ be all elements of linear non-associative algebra under 2-element field with mutiplication table of basic elements: $1a=a1=a, 1b=b1=b, ab=ba=0.$
For this "anything" we have $(a,b)^\top (a,b)=I$
PS. But I see no point in introducing multiplication of matrices for such "anything"