Please verrify for correctness. And I don't know where the assumption of having characteristic $\not = 2$ is used, please comment.
Lemma: Let $F$ be a field with characteristic $\not = 2$, $a, b \in F$, and $\sqrt{a},\sqrt{b} \not \in F$. Then $$F(\sqrt{a}) = F(\sqrt{b}) \iff \sqrt{a b} \in F.$$
Proof: $``\leftarrow"$ Use $\sqrt{b} = \sqrt{a} \sqrt{ab}\frac{1}{a}.$
$``\rightarrow"$ $\text{Gal}(F(\sqrt{a})/F) \overset{\sim}{\rightarrow} \mathbb{Z}/2\mathbb{Z}.$ We have that $\text{Gal}(F(\sqrt{a})/F) = \{\text{Id}, \sigma \}$, with $\sigma$ the conjugation. We have the basis $\{1, \sqrt{a}\}$ for $F(\sqrt{a})/F$, hence $F(\sqrt{a}) = \{x+y\sqrt{a}\; |x,y \in F \}$, with the conjugation acting as $x+y\sqrt{a} \overset{\sigma}{\mapsto}x -y\sqrt{a}$. Hence we have that $\sigma(z) = z \iff z \in F$.
So we have $\sigma(\sqrt{a}) = - \sqrt{a}$.
Now by taking a similar route, and using the assumption, we can also have $\{1, \sqrt{b} \}$ as a basis for $F(\sqrt{a})$, and we would also have that $\sigma$ acts like $x+y\sqrt{b} \overset{\sigma}{\mapsto} x - y\sqrt{b}$.
So we have $\sigma(\sqrt{b}) = - \sqrt{b}$.
Finally: $\sigma(\sqrt{a b}) = \sqrt{ab}$.
Check for correctness
You need to be careful with $\frac1a$ when $a=0$.
Where you assumed characteristic $2$
The notation $\operatorname{Gal}(F(\sqrt a)/F)$ assumes that $F(\sqrt a)/F$ is separable, but $X^2 - a$ has derivative $0$ in characteristic $2$.
Also, in characteristic $2$ we have $\sqrt a = -\sqrt a$, so your $\sigma$ is just the identity, and it certainly is not true that $\sigma(z) = z \iff z \in F$.
Counterexample in characteristic $2$
Set $F = \Bbb F_2(t)$, then $F(\sqrt t) = F(\sqrt{t+1})$ but $\sqrt{t(t+1)} \notin F$