There are necessarily further operators: a von Neumann algebra of operators if, as I am assuming, your representations are made of unitary operators.
All those operators can be constructed out of the orthogonal projectors commuting with the representation as complex linear combinations of them, since a complex von Neumann algebra is the closure of the complex span of its orthogonal projectors and here the closure is not necessary since the dimension is finite.
As a matter of fact you therefore should look for the invariant subspaces of each subrepresentation $R_i \otimes R_j$ in the corresponding product space $\cal{H}_i \otimes {\cal H}_j$ and the span of the orthogonal projectors onto these subspaces gives rise to the full space of operators commuting with $R_i \otimes R_j$. In particular, reversing the procedure, with $S_{jj}$ you can construct the subspace of symmetric and the subspace of antisymmetric tensors in ${\cal H}_j \otimes {\cal H}_j$, the projectors are $(I+S_{jj})/2$ and $(I-S_{jj})/2$, all complex combinations of these operators commute with the representation.
The fact that there are however many subspaces is already evident if looking at the case of $SU(2)$. In this case there is a canonical decomposition of $R_i\otimes R_j$ and $\cal{H}_i \otimes {\cal H}_j$ into orthogonal irreducible representations and the complex span of the corresponding projectors define operators in the commutant of $R_i\otimes R_j$ which actually exhaust this commutant (see below).
The said decomposition of $R_i\otimes R_j$ is nothing but the standard Clebsh-Gordan decomposition. In that case the procedure is relatively easy since the irreducible representations of $SU(2)$ are completely fixed by the eigenvalue of the unique Casimir operator $J^2$ and the overall argument reduces to look for the single eigenvalues of $J^2$ admitted for a product $R^{(j)}\otimes R^{(j')}$ of a pair of irreducible representations of $SU(2)$. Each eigenvalue defines an invariant subspace and the complex linear combinations of the corresponding orthogonal projectors are in the commutant of the representation in ${\cal H}^{(j)}\otimes {\cal H}^{(j)'}$.
Here you find the analogous decomposition for $SU(3)$, for $n>3$... I do not know, sorry :). You could try to have a look at Barut-Raczka's book on representations of groups.
An interesting fact is the following.
If the irreducible representations of a compact group are exactly all common eigenspaces of (mutually commuting) Casimir operators and each set of eigenvalues appears at most once, then the operators which commute with a tensor product of two irreducible representations $R_i \otimes R_j$ (a) are mutually commuting and (b) they are exactly the complex combinations of the orthogonal projectors onto the said common eigenspaces.
In fact: the space $\cal{H}_i \otimes {\cal H}_j$ is direct sum of mutually orthogonal irreducible representations by Peter-Weyl theorem which are eigenspaces of the Casimirs. If an operator $A$ commutes with the product representation $R_i \otimes R_j$ it also commute with its Casimir operators and thus it leaves invariant their common eigenspaces, i.e., the spaces of the irreducible representations that decompose $\cal{H}_i \otimes {\cal H}_j= \sum_k {\cal H}^{(i,j)}_k$. Since every such subspace ${\cal H}^{(i,j)}_k$ is irreducible, Schur's lemma entails that $A$ has the form $a_k I_k$ in every irreducible subspace, where $I_k$ is the identity operator on ${\cal H}^{(i,j)}_k$, and $a_k \in \mathbb C$. In summary the von Neumann algebra of the operators commuting with $R_i\otimes R_j$ is just made of the operators of form $\sum_k a_kI_k$, for some compex numbers $a_k$. This algebra is evidently Abelian.
