The numerical instability the code is meant to avoid is not one that occurs only if $\cos\theta$ is calculated to be greater than $1$ due to rounding; it occurs already if $\cos\theta$ is close to $1$, since the exact formula uses division by $\sin\theta=\sqrt{1-\cos^2\theta}$. So the answer is, no, you shouldn't replace that line; it's there for a good reason. You can experiment with increasing the bound to see whether it's been optimally chosen, but you shouldn't replace it by $1$.
There is a definitely a connection, but how good of a connection depends on your expectations. There is a lot of sexy interplay between the metric geometry using Clifford algebras and the quaternions. Clifford (or "geometric") algebras encode geometric information about the underlying space.
I'd like to mention three such algebras. First of all, $C\ell_{0,2}(\mathbb{R})=\mathbb{H}$.
Secondly, for ordinary Euclidean $\mathbb{R}^3$, the algebra $C\ell_{3,0}(\mathbb{R})$ contains $\mathbb{H}$ as a subalgebra of elements which enact rotations in the traditional quaternion way. (Of course, LOTS of Clifford algebras contain copies $\mathbb{H}$ but this is the one that does it in the "natural way".) These are the so-called rotors in the algebra.
Lastly, moving up to Minkowski space with $C\ell_{1,3}(\mathbb{R})$, the rotors are a bit different than the quaternions. The encompass both spatial rotations and Lorentz boosts and mixtures of the two. I think the phenomenon you observed might be a quaternion-like shadow of these rotors.
In your post it looked like you were reading a paper using biquaternions (complex quaternions), and I can recommend another good paper on that if you haven't found it already: check out Lambek's If Hamilton Had Prevailed: Quaternions in Physics (1995) Math.Intelligencer. He is generally known for good exposition and he's knowledgable about this topic in particular (it was part of his dissertation).
P.S.: If you're wondering about $C\ell_{3,1}(\mathbb{R})$, it's also interesting, but it is nonisomorphic to $C\ell_{1,3}(\mathbb{R})$ as $\mathbb{R}$ algebras. For the purposes of physics, I don't think any evidence has arisen to choose one as "better".
Best Answer
The key challenge is choosing a field for $\Bbb H$ to be a space over. (Bear in mind $\Bbb H$, unlike fields, isn't commutative.) One way to do this is to consider $\Bbb H$ a $2$-dimensional Hilbert space over $\Bbb C$ with basis $1,\,j$, so $a+bi+cj+dk=(a+bi)1+(c+di)j$ is a unique decomposition. Then $$\langle a+bi+cj+dk,\,e+fi+gj+hk\rangle:=(a-bi)(e+fi)+(c-di)(g+hi)$$is a suitable inner product. In particular$$\langle a+bi+cj+dk,\,a+bi+cj+dk\rangle=(a-bi)(a+bi)+(c-di)(c+di)=a^2+b^2+c^2+d^2,$$as expected. What we can't do, however, is use the ordinary quaternion multiplication $q_1^\ast q_2$, in which $q_1$ is conjugated, as an inner product, because IPs have to live in the field, in this case $\Bbb C$.