Let $\mathcal{R}$ be representations of $U(\mathfrak{g})$ and $\mathcal{R}_q$ be type 1 representations of $U_q(\mathfrak{g})$. As you mentioned, $\mathcal{R}$ and $\mathcal{R}_q$ are equivalent as categories. This is because they are both semisimple and they have the same set of irreducibles. Moreover, $\mathcal{R}$ and $\mathcal{R}_q$ have the same fusion ring. You can see this by considering characters. They are not equivalent as tensor categories because the associators are fundamentally different. You can see this concretely in the case $\mathfrak{g} = \mathfrak{sl}_2$ by computing 6j symbols as described in the book The classical and quantum 6j symbols.

More abstractly, the 6j symbols are "coordinates" on the moduli stack of tensor categories with a fixed fusion ring. For example, if $G$ is a finite group, then the moduli stack of tensor categories with fusion ring $\mathbb{C}[G]$ is $H^3(G,\mathbb{C}^{\times})$.

Let $\mathcal{R}$ be representations of $U(\mathfrak{g})$ and $\mathcal{R}_q$ be type 1 representations of $U_q(\mathfrak{g})$. As you mentioned, $\mathcal{R}$ and $\mathcal{R}_q$ are equivalent as categories. This is because they are both semisimple and they have the same set of irreducibles. Moreover, $\mathcal{R}$ and $\mathcal{R}_q$ have the same fusion ring. You can see this by considering characters. They are not equivalent as tensor categories because the associators are fundamentally different. You can see this concretely in the case $\mathfrak{g} = \mathfrak{sl}_2$ by computing 6j symbols as described in the book The classical and quantum 6j symbols.

More abstractly, the 6j symbols are "coordinates" on the moduli stack of tensor categories with a fixed fusion ring. For example, if $G$ is a finite group, then the moduli stack of tensor categories with fusion ring $\mathbb{C}[G]$ is $H^3(G,\mathbb{C}^{\times}) / {\rm Out}(G)$.

Let $\mathcal{R}$ be representations of $U(\mathfrak{g})$ and $\mathcal{R}_q$ be type 1 representations of $U_q(\mathfrak{g})$. As you mentioned, $\mathcal{R}$ and $\mathcal{R}_q$ are equivalent as categories. This is because they are both semisimple and they have the same set of irreducibles. Moreover, $\mathcal{R}$ and $\mathcal{R}_q$ have the same fusion ring. You can see this by considering characters. They are not equivalent as tensor categories because the Associators are fundamentally different. You can see this concretely in the case $\mathfrak{g} = \mathfrak{sl}_2$ by computing 6j symbols as described in the book The classical and quantum 6j symbols.

More abstractly, the 6j symbols are "coordinates" on the moduli stack of tensor categories with a fixed fusion ring. For example, if $G$ is a finite group, then the moduli stack of tensor categories with fusion ring $\mathbb{C}[G]$ is $H^3(G,\mathbb{C}^{\times})$.

Let $\mathcal{R}$ be representations of $U(\mathfrak{g})$ and $\mathcal{R}_q$ be type 1 representations of $U_q(\mathfrak{g})$. As you mentioned, $\mathcal{R}$ and $\mathcal{R}_q$ are equivalent as categories. This is because they are both semisimple and they have the same set of irreducibles. Moreover, $\mathcal{R}$ and $\mathcal{R}_q$ have the same fusion ring. You can see this by considering characters. They are not equivalent as tensor categories because the associators are fundamentally different. You can see this concretely in the case $\mathfrak{g} = \mathfrak{sl}_2$ by computing 6j symbols as described in the book The classical and quantum 6j symbols.

More abstractly, the 6j symbols are "coordinates" on the moduli stack of tensor categories with a fixed fusion ring. For example, if $G$ is a finite group, then the moduli stack of tensor categories with fusion ring $\mathbb{C}[G]$ is $H^3(G,\mathbb{C}^{\times})$.