Antipode of Hopf algebra in braided monoidal category is an algebra antihomomorphism?

This idea should work and Majid's proof uses this. I think your reasoning illustrates how the proof is related to the one in the (non braided) Hopf algebra case.

You can verify that $\mu\circ \sigma\circ(S\otimes S)$ indeed is a right convolution inverse of $\mu$ as follows. First, by naturality of $\sigma$ twice you get \begin{align*} &\mu \circ (\mu \otimes \mu) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \operatorname{id} \otimes S \otimes S) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ (\Delta \otimes \Delta)\\ &=\mu \circ (\mu \otimes \mu(S\otimes\operatorname{id})) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ ((\operatorname{id}\otimes S)\Delta \otimes \Delta). \end{align*} Then apply naturality of braiding (after using the Yang-Baxter equation of the braiding) to $\Delta$, yielding that the above equals (if you also use associativity of multiplication twice) \begin{align*} \mu\circ(\mu\otimes \operatorname{id})\circ(\operatorname{id}\otimes \mu(\operatorname{id}\otimes S)\Delta\otimes \operatorname{id})\circ(\operatorname{id}\otimes \sigma)\circ((\operatorname{id}\otimes S)\Delta \otimes \Delta). \end{align*} Here, you can find the term $\mu(\operatorname{id}\otimes S)\Delta$ which equals $1\circ\varepsilon$ and simplifies everything (under use of the axioms of the unit), to $$(\mu(\operatorname{id}\otimes S)\Delta)\otimes \varepsilon$$ which now gives $1\circ(\varepsilon\otimes \varepsilon)$.

You need to convince yourself that if a map has a right and left convolution inverse, they are equal. The same proof of this fact also works in a braided monoidal category as it only uses (co)associativity. Note that this does not require that $H\otimes H$ is a Hopf algebra (which it is not in general). However, $H\otimes H$ is both a coalgebra and algebra and this is all we need.

The above reasoning will again be more clear to me when using graphical calculus. Without it, I would have a hard time coming up with the steps necessary:

(Note that the functional identities above do repress $\circ$ when it appears within tensor product legs)