Even though $f$ is qcqs and $S$ is qcqs, the natural morphism $j^*f_{\#}\to f_{\#}'{j'}^*$ is not an isomorphism. Here is an example:
Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$ and $f = \mathrm{id}_S$. Then, $f_{\#}$ is exactly the coherator $Q_S:\mathsf{Mod}(\mathcal{O}_S)\to \mathsf{QCoh}(S)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$.
Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$. Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$ (cf. Thomason-Trobaugh, Appendix B. 14).
By using this observation, we can conclude the following:
Lemma. Let $F$ be an $\mathcal{O}_S$-module such that for any $f\in A$, if we write $S_f :\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$, then the natural morphism $Q_S(F)|_{S_f}\to Q_{S_f}(F|_{S_f})$ is an isomorphism.
Then, $F$ is a quasi-coherent sheaf.
Since $Q_S(F)$ is quasi-coherent, it holds that $\Gamma(S_f, Q_S(F)|_{S_f}) = \Gamma(S, Q_S(F))_f$. Since $Q_S(F) = \Gamma(S, F)\tilde{ \ }$, it holds that $\Gamma(S, Q_S(F)) = \Gamma(S, F)$. Hence, it holds that $\Gamma(S_f, Q_S(F)|_{S_f}) = \Gamma(S, F)_f$. Thus, it follows from the assumption that the natural morphism $\Gamma(S,F)_f\to \Gamma(S_f, F)$ is an isomorphism.
Since the family of open subschemes $\left\{ S_f \,\middle|\, f\in A\right\}$ is an open base of $S$, this implies that for any $x\in S$, the morphism $Q_S(F)_x\to F_x$ induced by taking stalks of the natural morphism $Q_S(F)\to F$ is an isomorphism. Hence, in particular, $Q_S(F)\to F$ is an isomorphism of sheaves.
This implies that $F$ is a quasi-coherent sheaf. ◻︎
By the above Lemma, we can conclude that for any non-quasi-coherent $\mathcal{O}_S$-module $F$, there exists an element $f\in A$ such that the natural morphism $Q_S(F)|_{S_f}\to Q_{S_f}(F|_{S_f})$ is not an isomorphism. In particular, if we write $j: S_f \hookrightarrow S$ for the natural open immersion, then the natural morphism $j^*\mathrm{id}_{S,\#}F \to \mathrm{id}_{S_f, \#}{j'}^*F$ is not an isomorphism.
A more concrete counterexample:
Let $A$ be a DVR. Write $K$ for the field of fractions of $A$. Define an $\mathcal{O}_S$-module $F$ as follows: $F(\mathrm{Spec}(A)) = 0, F(\mathrm{Spec}(K)) = K$.
Then, $Q_{\mathrm{Spec}(A)}(F) = 0$, $Q_{\mathrm{Spec}(K)}(F|_{\mathrm{Spec}(K)}) = \tilde{K}$, and, moreover, the natural morphism $0\to K$ is not an isomorphism.
Note that if $g:T\to S$ is a morphism of schemes, then the scheme-theoretic image of $g$ is equal to the closed subscheme corresponding to the quasi-coherent ideal sheaf $Q_S(\mathrm{ker}(g^{\#}:\mathcal{O}_S\to g_*\mathcal{O}_T))\subset \mathcal{O}_S$. The above $F$ is a special case of this: $T:\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}} \mathrm{Spec}(A/(\pi^i))$, where $\pi\in A$ is a uniformizer, and $g: T \to \mathrm{Spec}(A)$ for the canonical morphism of schemes. Then, $F = \mathrm{ker}(\mathcal{O}_{\mathrm{Spec}(A)}\to g_*\mathcal{O}_T)$, and the scheme theoretic image of $g$ is equal to $\mathrm{Spec}(A)$.