Then $F$ is fully faithful.
Proof.
Denote $F_ r$ and $F_ l$ the right and left adjoints of $F$. For $E \in S$ choose a distinguished triangle
\[ E \to F_ r(F(E)) \to C \to E[1] \]
where the first arrow is the unit of the adjunction. For $E' \in S$ we have
\[ \mathop{\mathrm{Hom}}\nolimits (E', F_ r(F(E))[i]) = \mathop{\mathrm{Hom}}\nolimits (F(E'), F(E)[i]) = \mathop{\mathrm{Hom}}\nolimits (E', E[i]) \]
The last equality holds by assumption (4). Hence applying the homological functor $\mathop{\mathrm{Hom}}\nolimits (E', -)$ (Derived Categories, Lemma 13.4.2) to the distinguished triangle above we conclude that $\mathop{\mathrm{Hom}}\nolimits (E', C[i]) = 0$ for all $i \in \mathbf{Z}$ and $E' \in S$. By assumption (2) we conclude that $C = 0$ and $E = F_ r(F(E))$.
For $K \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ choose a distinguished triangle
\[ F_ l(F(K)) \to K \to C \to F_ l(F(K))[1] \]
where the first arrow is the counit of the adjunction. For $E \in S$ we have
\[ \mathop{\mathrm{Hom}}\nolimits (F_ l(F(K)), E[i]) = \mathop{\mathrm{Hom}}\nolimits (F(K), F(E)[i]) = \mathop{\mathrm{Hom}}\nolimits (K, F_ r(F(E))[i]) = \mathop{\mathrm{Hom}}\nolimits (K, E[i]) \]
where the last equality holds by the result of the first paragraph. Thus we conclude as before that $\mathop{\mathrm{Hom}}\nolimits (C, E[i]) = 0$ for all $E \in S$ and $i \in \mathbf{Z}$. Hence $C = 0$ by assumption (3). Thus $F$ is fully faithful by Categories, Lemma 4.24.4.
$\square$
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