Although this question arises from an exercise, you have already done almost all of the work to solve it. Your question is basically "should I be using the energy $E$ or the average energy $\overline{E}$ here?" or "am I justified in solving the problem this way?". This is a conceptual question, connected with the transformation of ensembles between microcanonical and canonical, which I believe this exercise is trying to help illustrate. So this answer will concentrate on that context.
You are correct that the energy $E$ of your paramagnet system is a simple linear function of the spin excess $2s$. So, go ahead, as you imply in your question: use that equation to substitute for $s$ in your entropy equation, to give you an entropy function $S(E)$. Use that to calculate a Helmholtz function, at a given $T$, which for the moment I'll call $\mathcal{F}(E)=E-TS(E)$. Do as the exercise asks: differentiate $\mathcal{F}(E)$ to find the energy $E=\hat{E}$ which minimizes $\mathcal{F}(E)$ and compare with the average energy $\overline{E}$ which comes from analyzing this model in the canonical ensemble (at temperature $T$). I'm leaving all of that to you.
The conversion between microcanonical and canonical ensembles is found in many statistical mechanics books. In the canonical ensemble at temperature $T$, the probability distribution function for energy may be written
$$
\mathcal{P}(E) \propto \Omega(E)\exp(-E/k_BT)
$$
where $\Omega(E)$ is the density of states (number of states per unit energy),
and $\exp(-E/k_BT)$ is the Boltzmann factor for energy $E$. NB, here I am glossing over the fact that $E$ only takes discrete values in this spin model: I am treating it as a continuous variable (as the exercise expects you to do, when you differentiate with respect to $E$).
$\Omega(E)$ is a very rapidly increasing function of $E$, for large $N$. The Boltzmann factor is a very rapidly decreasing function of $E$ (which is, after all, an extensive variable). The consequence is that $\mathcal{P}(E)$ has a very sharp peak at some value $E=\hat{E}$. Moreover, because of the sharpness, this value will be very close to the average energy $\overline{E}$. Before differentiating to find the maximum of $\mathcal{P}(E)$, it is convenient to take logs:
$$
-k_BT \ln \mathcal{P}(E) = E- k_BT\ln\Omega(E) + \text{const} = E-TS(E) + \text{const} = \mathcal{F}(E) + \text{const},
$$
where we recognize $S(E)=k_B\ln\Omega(E)$,
the formula which was used to obtain your entropy expression.
Differentiating $\mathcal{F}(E)$ with respect to $E$,
setting to zero,
and hence finding $\hat{E}$,
is the key step.
In the process of doing that, one gets to match up the energy $E$
of the microcanonical ensemble with the temperature of the canonical ensemble,
by requiring
$$
\left . \frac{\partial S(E)}{\partial E} \right|_{E=\hat{E}}=\frac{1}{T}
$$
One can go further and discuss the width of the $\mathcal{P}(E)$ distribution,
the link between $\mathcal{F}(\hat{E})$ and the thermodynamic free energy $F$,
and so on,
but that would go beyond the scope of the question.
The main point is that,
provided $N$ is large,
we get the desired result:
the temperature $T$ in the canonical ensemble is chosen to make
$\overline{E}\approx \hat{E}=E$ in the microcanonical ensemble.
