*Partial answer*  

The second fusion ring admits no *unitary* categorification. It is proved in [this paper][1], using Character Table and Fourier Analysis. Here are some explanations:  

Let $\mathcal{F}$ be a commutative fusion ring. Then its fusion matrices $(M_i)$ are commuting and normal so simultaneously diagonalisable: there is an invertible matrix $P$ such that $P^{-1}M_iP = diag(\lambda_{i,j})$. Let us call $\Lambda=(\lambda_{i,j})$ the *character table* of $\mathcal{F}$. Note that if $G$ is a finite group and $\mathcal{F}$ the Grothendieck ring of $Rep(G)$ then $\Lambda$ is the usual character table of $G$.  

If $\mathcal{F}$ is the Grothendieck ring of a unitary fusion category, and if we choose $\lambda_{i,1}$ to be $\Vert M_i \Vert$, then by Corollary 7.5 in [this paper][1], for every triple $(j,k,\ell)$ $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$
Note that for $Rep(G)$ this identity admits a combinatorial/probabilistic interpretation (see [here][2]).


Now observe the character table of the second fusion ring above, in particular its last column, then
$$ \frac{1^3}{1} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{0^3}{5} + \frac{1^3}{6} + \frac{(-3)^3}{7} + \frac{2^3}{7} =  -\frac{65}{42}<0.$$
It follows that the second fusion ring above admits no unitary categorification.  
 
Note that the identity holds for the first fusion ring.

  [1]: https://arxiv.org/abs/1910.12059v1
  [2]: https://mathoverflow.net/q/344968/34538