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In "$\Rightarrow$" you assume there is a basis with respect to which $X$ is diagonal and $Y$ is a monomial matrix. This is stronger than "simultaneously monomializable." Indeed, take $X,Y$ to be the $3\times 3$ permutation matrices corresponding to $(1\/2)$ and $(1\/2\/3)$. Then they are monomial in the standard basis, but $YXY^-1$ corresponds to $(2\/3)$, which does not commute with $(1\/2)$.
The braid group $\mathcal{B}_d$ generators $\sigma_i$ act on any $\mathcal{H}_q(d)$ module by an operator with (at most) two eigenvectors. So if $V$ and $W$ are two $\mathcal{H}_q(d)$ irreps and $\mathcal{H}_q(d)$ had a bialgebra structure then the braid group would act on $V\otimes W$. In the case of $q=1$ this $\B_d$ action is diagonal, but in the $q$ generic (or root of unity) case the diagonal action would give us the wrong eigenvalues for $\sigma_i$, assuming a bialgebra structure. This is not a proof that such a bialgebra structure does not exist, but it does seem problematic..
Various German algebraists (Noether for example) used Fraktur script in their work at least back to the 1910s. So one reasonable guess would be Hermann Weyl, who advanced the theory of Lie algebras in the 1930s.
I would add to Vendramin's answer that for finite-dimensional semisimple pointed Hopf algebras $H$, the representation category $Rep(H)$ is much like a group: all the simple objects (irreps.) are invertible. As such these are essentially classified, and the \emph{non-semisimple} case is of greatest interest.
