Working from the definition that $\mathbf a \times \mathbf b =\|\mathbf a\|\|\mathbf b||\sin(\theta)$ I want to know why there are only two possible outcomes in a cartesian system. Since we already have that it's distributive and anticommutative, we can start by defining the possible outcomes of the product in its cartesian form.
$$\mathbf i\times \mathbf j=\pm \mathbf k \\ \mathbf j\times \mathbf k=\pm \mathbf i \\ \mathbf k\times \mathbf i=\pm \mathbf j$$
It appears there are actually 8 options. I know intuitively that this can't be the case or the cross product wouldn't be continuous over $\mathbf i,\mathbf j,\mathbf k$. The problem is I can't show it rigorously from first principles. (Without simply invoking that it has to be continuous and using counterexamples)
There are 8 options, but there are only two indistinguishable families of the product.
Suppose we start with $i\times j = k, j\times k = i, k\times i= j$, swapping any orthogonal vector for its negative counterpart will be exactly equivalent to swapping the order of multiplication of two of the cross products. This gives us 4 of the options. Any further sign changes will only convert the cross product into one of these 4 variations. Hence choosing any of these 4 options is equivalent to redefining the coordinate system.
Now let's start with $j\times i = k, k\times j = i, i\times k= j$ or by the anticommutative law, $i\times j = -k, j\times k = -i, k\times i= -j$. Now changing the sign of any basis vector will allow us to convert between the remaining 4 cross products, but we cannot switch to the previous definition.
It is much easier to prove the distributive property of the cross product when we start with the cartesian definition. This argument applies just as well for both methods