I realize that the question posed in the title has already been addressed here: Identities that connect antipode with multiplication and comultiplication, where the graphical calculus proof provided by Majid is cited. However, I am wondering if there is a more algebraic proof that can be formulated without the use of graphical calculus.

If we have a Hopf algebra $(H, \mu, \eta, \Delta, \varepsilon, S)$ in the category of vector spaces, where $\mu$ is the multiplication map, $\eta$ is the unit map, $\Delta$ is the comultiplication, $\varepsilon$ is the counit, and $S$ is the antipode, then one can show that $S \circ \mu$ is a left convolution inverse of $\mu$ and that $\mu \circ \tau \circ (S \otimes S)$ is a right convolution inverse of $\mu$ (where $\tau$ is the flip map). From here, then $S \circ \mu = \mu \circ \tau \circ (S \otimes S)$.

I am aiming to replicate the above proof for an arbitrary braided monoidal category, replacing the flip map $\tau$ with a braiding $\sigma$. Showing that $S \circ \mu$ is a left convolution inverse of $\mu$ is easy, just using that the comultiplication $\Delta$ is an algebra homomorphism. I can not seem to get that $\mu \circ \sigma \circ (S \otimes S)$ is a right convolution inverse of $\mu$, however. We should have \begin{align*} &\mu \ast (\mu \circ \sigma \circ (S \otimes S))\\ &= \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) \end{align*} but I can not see how to get anywhere with this, particularly trying to figure out what to do with the map $\operatorname{id} \otimes \operatorname{id} \otimes \sigma$. It seems as though the antipode equation for $H \otimes H$ should be used, but I am not quite sure how to implement it. Any ideas would be appreciated.