Let $\mathbf{A}=(A,\odot,\mathbf{1}_A, ^*)$ be a (not necessarily associative) unital $*$-algebra over a real closed field $(K,\preceq)$. I shall say that $\mathbf{A}$ is $\textit{nicely normed}$ provided
$$\mathbf{a}+\mathbf{a}^{*} \in K[\mathbf{1}_A]$$
for all $\mathbf{a} \in A$, and additionally $\mathbf{a}\odot\mathbf{a}^{*} \succ \mathbf{0}_A$ for all nonzero $\mathbf{a}$. Here $K[\mathbf{1}_A]$ denotes the $K$-subalgebra generated by $\mathbf{1}_A$, naturally inheriting the ordered field structure from $K$.
Does the map $\mathbf{a} \mapsto \sqrt{\mathbf{a}\odot \mathbf{a}^{*}}$ define a submultiplicative "norm" $\| \cdot \|:\mathbf{A} \to K[\mathbf{1}_A] \cong K$? If not, what are the minimal constraints on $\mathbf{A}$ and $K$ for it to be so?
One can easily see that $\| \lambda \cdot \mathbf{a}\|=|\lambda| \cdot \| \mathbf{a} \|$, and $\| \cdot \|$ is clearly non-degenerate, so my main focus is getting subadditivity (i.e. the triangle inequality) and submultiplicativity to hold.
$\textbf{Edit}$: It may be useful to note that the map $(\mathbf{a},\mathbf{b}) \mapsto \frac{1}{2}(\mathbf{a}\odot \mathbf{b}^{*} + \mathbf{b} \odot \mathbf{a}^{*})$ defines a non-degenerate symmetric $K$-bilinear form, and hence that $\mathbf{a} \mapsto \mathbf{a}\odot \mathbf{a}^{*}$ defines an anisotropic quadratic form over $K$.
These assumptions imply that
$$\langle a, b \rangle = \frac{a^{\ast} b + b^{\ast} a}{2} \in K$$
defines a $K$-bilinear inner product on $A$ with values in $K$, so $\sqrt{ \langle a, a \rangle } = \sqrt{ a^{\ast} a }$ is a Euclidean / Hilbert space norm (and so in particular satisfies the triangle inequality). There is no need to assume that $K$ is rigid. Also the first assumption implies the second so the second assumption is redundant.
Moreover the norm is straightforwardly multiplicative. We have
$$(ab)^{\ast} ab = b^{\ast} a^{\ast} ab = a^{\ast} a b^{\ast} b$$
because by assumption $a^{\ast} a \in K$ is a scalar and so is central, hence can be commuted past $b^{\ast}$.