Suppose that $\mathbb{R}^n$ is a unital division algebra. That is, $\mathbb{R}^n$ is furnished with a bilinear map $$ *: \mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}^n $$ such that $$ x*y = 0 \implies x=0 \text{ or }y=0. $$ In addition, there is a distinguished element $\mathbf{1}\in \mathbb{R}^n$ where $$ \mathbf{1}*x = x*\mathbf{1} = x $$ for all $x\in\mathbb{R}^n$.
Is it now true that $$ \| \mathbf{1} \| =1 $$ where $\|\cdot\|$ is the usual euclidean norm on $\mathbb{R}^n$? In other words, is the unit of $\mathbb{R}^n$ in the unit sphere $S^{n-1}$?
I have tried proving that $$ \|x*y\| =\|x\|\|y\| \quad \forall x,y\in \mathbb{R}^n $$ without success. Of course, if this above equation holds, then $\|\mathbf{1}\|=1$ is an immediate consequence.
A positive answer to this question would help me prove the following assertion: If $\mathbb{R}^n$ is a unitial division algebra, then the sphere $S^{n-1}$ is an $H$-space.
Of course, I cannot use the classification of division algebras over $\mathbb{R}$, to just say "well in all four cases, $\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}$ the unit clearly has norm $1$."
EDIT: Apparently the result $\|\mathbf{1}\|=1$ is false, which puts me back at square one.
You don't need the fact you're trying to prove to prove this. $S^{n-1}$ can be defined without any reference to the Euclidean norm, as the quotient of $\mathbb{R}^n \setminus \{ 0 \}$ by scalar multiplication by positive reals (very similar to the construction of real projective space). The multiplication on an $n$-dimensional unital division algebra induces a multiplication on this quotient because, by hypothesis, the product of two nonzero elements is nonzero, and the multiplication is bilinear so respects scalars. And the image of $1$ under this quotient map continues to act as an identity.
Said more concretely, instead of using the multiplication $\ast$ you started with, you can consider "normalized multiplication" $\frac{x \ast y}{\| x \ast y \|}$, which by construction has the property that if $\| x \| = 1$ and $\| y \| = 1$ then $ \| x \ast y \| = 1$, and also that $1$ continues to act as an identity (on the unit sphere).
The recognition that we can scale multiplication arbitrarily (in a way that depends on the vector, even) shows that the fact you're trying to prove can't possibly be true, since e.g. we can consider $\mathbb{R}$ with the scaled multiplication $x \ast y = 2xy$, which has an identity given by $\frac{1}{2}$ which does not have Euclidean norm $1$.