If $3 \not \in S$ then the answer to Question($S$) is **yes**.  

There is an integral commutative fusion ring which is not of Frobenius type. So we are reduced to $3 \in S$, except if the fusion ring below is one among few anomalies (which then require to be classified).  
Let $[d(b_1), d(b_2), \dots , d(b_r)]$ be the type of the fusion ring.

*Here is an example*: rank $4$, global dimension $15$, type $[1,1,2,3]$, fusion rules:  

 $$\left(\begin{smallmatrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1
\end{smallmatrix} \right) , \ \left(\begin{smallmatrix}
0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1
\end{smallmatrix} \right) , \ \left(\begin{smallmatrix}
0&0&1&0\\0&0&1&0\\1&1&1&0\\0&0&0&2
\end{smallmatrix} \right) , \ \left(\begin{smallmatrix}
0&0&0&1\\0&0&0&1\\0&0&0&2\\1&1&2&1
\end{smallmatrix} \right) $$

and character table:  
$$\left[\begin{matrix} 1&1&1&1\\1&1&-1&1\\2&-1&0&2\\-2&0&0&3  \end{matrix} \right]$$

It is not of Frobenius type because $2$ does not divide $15$.