Let's take a modified version of the axioms above. We want a coordinate system for 3D rotations with the following features:

1. The coordinates form a set equal to $\mathbb R^n$ for some $n$.
2. All non-zero coordinates represent some rotation.
3. Every rotation is represented by some coordinate.
4. If $v$ represents a 3D rotation, then any $\lambda v$ where $\lambda$ is a non-zero scalar, represents the same rotation. The motivation for this axiom is that we want to find a projective coordinate system for 3D rotations.
5. The composition of rotations corresponds to a bilinear product between coordinates. In particular, such a product consists of only the operations $+, \times, -$.

Let $\mathbf 1$ be the coordinate for a null rotation, $i$ be the coordinate for a half-turn around the $\vec i$ vector, and $j$ and $k$ be likewise for the $\vec j$ and $\vec k$ vectors. This all follows from axiom 3.

Let's consider the point $i$. Immediately, it's clear that $i^2 = \lambda \mathbf 1$ for some scalar $\lambda$. This scalar could be anything at this point except $0$.

By axiom (1) and (2), every non-zero linear combination of $\mathbf 1$ and $i$ represents some rotation. It is weird to take linear combinations of projective coordinate, but it is certainly possible, and it allows us to narrow down the space of possibilities. We see that these linear combinations generate a 2D algebra. It turns out there are only three 2D algebras up to isomorphism: The complex numbers, the dual numbers, and the split-complex numbers. We see that up to isomorphism, the algebra we've generated must be the complex numbers as the other ones contain zero divisors, and all rotations are invertible. Alternatively, we may argue using the orders of the elements: Rotations contain elements of any order.

The subalgebra generated by $\{\mathbf 1, i\}$ must correspond to some continuous subgroup of the 3D rotations. The only such subgroups that exist are rotations about some axes. Since $i$ represents a rotation about the $\vec i$ axis by a half turn, the group we've generated must be the set of all rotations about $\vec i$.

A similar argument applies to $j$ and $k$.

Finally, what is $ij$? Well, for geometric reasons it must be $\lambda_{ij} k$ for some scalar $\lambda_{ij}$. Moreover, we have $(iji^{-1})^2 = ij^2i^{-1}=-1$; so $\lambda_{ij}=\pm 1$. By geometric considerations also, $ji = \lambda_{ji}k$, and we can likewise conclude that ${ji} = \pm k$. Finally, we conclude that $i$ and $j$ anti-commute by starting with $(ij)^2 = -1$ and rearranging to get $ij = -(ij)^{-1}=-j^{-1}i^{-1}=-ji$.

What's the motivation?

Essentially, I posit that a transformation group can often be endowed with a coordinate system or representation. This coordinate system is either cartesian or projective.

A Cartesian coordinate system for a transformation group is one where (except for a negligible subset) all elements of $\mathbb R^n$ for some $n$ are in one-to-one correspondence with that transformation group. We insist that the product be bilinear. The rotation-dilation group can be represented by the Cartesian coordinate system $\mathbb C$. The complex number $i$ now represents a quarter-turn.

In fact, the classic examples of Cartesian systems for transformation groups are rings of square matrices.

Now consider as an alternative projective coordinate systems for transformation groups. Here, a scalar multiple of a coordinate represents the same transformation. Again, we insist that almost all elements of $\mathbb R^n$ stand for some element of the group, and that the product be bilinear.

The classic example is the matrix representation of the group of affine transformations. Here, the matrices are specified up to a ratio. We are generalising this because we may want our coordinate system to be more specific for computational efficiency and accuracy reasons.

The complex numbers $\mathbb C$ are also a projective coordinate system for rotations (not dilations). But then $i$ represents a half-turn, not a quarter turn.

[Removed a subjective remark that's perhaps non-mathematical. If someone who's seen it thinks it's appropriate to include it here, then I might do that].