= Solution
For an invertible <braiding> $C$ satisfying the braid equation on $V\otimes V$, begin with the free associative <algebra over a field> on matrix coefficients $t_{ij}$, with $\Delta t_{ij}=\sum_kt_{ik}\otimes t_{kj}$ and $\varepsilon(t_{ij})=\delta_{ij}$. Impose precisely the coefficients of
$$
\delta_{V\otimes V}C=(C\otimes1)\delta_{V\otimes V},\qquad
\delta(e_i)=\sum_j e_j\otimes t_{ji}.
$$
This quotient is the <FRT bialgebra>, or <Faddeev-Reshetikhin-Takhtajan construction>. The relations form a <coideal> because composing two <coactions> that preserve $C$ again preserves $C$; the trivial <coaction> also preserves it, so the <counit> descends. Universality follows because any such <coaction> must satisfy exactly these coefficient relations.
For the given $C$, take $t_{11}=a$, $t_{12}=b$, $t_{21}=c$, $t_{22}=d$. Thus $\delta(e_1)=e_1\otimes a+e_2\otimes c$ and $\delta(e_2)=e_1\otimes b+e_2\otimes d$. The coefficient relations, after cancellation of $q^{-1/2}$, are
$$
\boxed{ba=qab,\ ca=qac,\ db=qbd,\ dc=qcd,\ bc=cb,\ da-ad=(q-q^{-1})bc.}
$$
These are exactly the defining relations of $M_q(2)$ in the inverse-parameter convention of Question 1, without setting the <quantum determinant> equal to one. They therefore define mutually inverse <algebra homomorphisms>: each named generator maps to the corresponding $t_{ij}$, and conversely. The matrix <comultiplication> and <counit> are the same, so the isomorphism is also one of <bialgebras>.
Here is the requested check of $ba=qab$ directly from $C$. Apply colinearity to $e_1\otimes e_2$ and compare the coefficient of $e_1\otimes e_1$. On the left, $C(e_1\otimes e_2)=q^{-1/2}e_2\otimes e_1$, giving $q^{-1/2}ba$. On the right, the corresponding term of $\delta(e_1\otimes e_2)$ is $e_1\otimes e_1\otimes ab$, and $C(e_1\otimes e_1)=q^{1/2}e_1\otimes e_1$. Hence $q^{-1/2}ba=q^{1/2}ab$, proving the relation. The other coefficient comparisons give the remaining displayed relations; no additional determinant relation is imposed in an <FRT bialgebra>.
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