Is it possible to implement the controlled-S gate, such that the inner gate between the CNOTs belongs to the Clifford?

It's well known how to implement a controlled-U gate when $$U$$ is a single qubit gate via the decomposition of $$U$$ into the product $$U=e^{i\phi}AXBXC$$ where $$ABC=I$$.

My question is: is it possible to implement the controlled-S gate in such a way that $$B$$ belongs to the Clifford group?

TL;DR: After sending $$D$$ across the equals sign, the right hand side is similar to a controlled Pauli operator and the left hand side is a diagonal operator. Their spectra turn out to be incompatible, so they cannot be equal.
If $$B$$ is Clifford, then $$BXB^\dagger=P$$ is a Pauli operator with eigenvalues $$\pm 1$$ and $$Q:=XP$$ is either $$\pm I$$ or a Pauli operator with eigenvalues $$\pm i$$. The former renders the circuit on the right hand side non-entangling, so assume the latter. Note that \begin{align} AXBXC&=AXPBC=AQA^\dagger\tag1 \end{align} so shifting the $$B$$ gate left to the position between $$C$$ and the first CNOT on the right hand side of the diagram in the question yields controlled-$$AQA^\dagger$$ followed by $$D\otimes I$$.
Moreover, by considering the action of the two sides of the diagram on the computational basis we see that $$D=\mathrm{diag}(d_1,d_2)$$ for some $$d_1,d_2\in\mathbb{C}$$. Now, hit both sides of the diagram with $$D^\dagger\otimes I$$. On the left hand side we obtain $$(D^\dagger\otimes I)\circ\text{CS}=\mathrm{diag}(\overline{d}_1, \overline{d}_1, \overline{d}_2, i\overline{d}_2).\tag2$$ On the right hand side we obtain an operator, similar to a controlled Pauli, with eigenvalues $$+1, +1, +i, -i$$. We conclude that the two operators cannot be equal if $$B$$ is Clifford.