So the answer is a bit surprising (maybe I have a mistake). You have an adjunction between $Fun(X,\mathrm{Sp})$ and $\mathrm{Sp}$ which in one direction sends a functor to its (homotopy) colimit and on the other hand sends a spectrum to the constant functor $X \to \mathrm{Sp}$ with that value. If $X$ has an $E_\infty$-structure then $Fun(X,\mathrm{Sp})$ inherits the Day convolution symmetric monoidal structure and the functor $colim: Fun(X,\mathrm{Sp}) \to \mathrm{Sp}$ is symmetric monoidal. As a result the right adjoint $const: \mathrm{Sp} \to Fun(X,\mathrm{Sp})$ carries a canonical lax-monoidal structure, and the adjunction $colim \dashv const$ can be promoted to an adjunction
$$Alg_{E_{\infty}}(Fun(X,\mathrm{Sp})) \rightleftarrows Alg_{E_{\infty}}(\mathrm{Sp})$$
on the level of $E_\infty$-algebras. In particular, if $E$ is an $E_\infty$-ring spectrum then $E_X = const(E)$ is naturally an $E_\infty$-algebra with respsect to the day convolution product. By adjunction one gets that
$$ Map_{Alg_{E_{\infty}}(Fun(X,\mathrm{Sp}))}(E_X,E'_X) \simeq Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(colim_XE_X,E') \simeq Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(\Sigma^{\infty}_+(X) \wedge E,E') $$
On the other hand, smash of $E_\infty$-ring spectra is just the $\infty$-categorical coproduct of ring spectra! We then get that
$$ Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(\Sigma^{\infty}_+(X) \wedge E,E') \simeq Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(\Sigma^{\infty}_+(X),E') \times Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(E,E') $$
In other words, to describe a map of $E_\infty$-algebras $E_X \to E'_X$ one needs to give a map of $E_\infty$-rings spectra $f:E \to E'$ together with a map of $E_\infty$-ring spectra $a:\Sigma^{\infty}_+(X) \to E'$. Alternatively, we may write a map of $E_\infty$-ring spectra $a:\Sigma^{\infty}_+(X) \to E'$ as a map of $E_{\infty}$-monoids $a:X \to Map(S^0,E')$. Given a pair $(f,a)$ the associated map $E_X \to E_X'$ is presumably given at the point $x \in X$ by the map $a_x\cdot f \in Map_{\mathrm{Sp}}(E,E')$.
Edit: Let $\mathcal{C} = Fun(X,\mathrm{Sp})$ with the Day convolution symmetric monoidal structure. The OP is asking if one can write $Map_{Alg_{E_\infty}(\mathcal{C})}(E_X,E'_X)$ as a mapping space of $E_\infty$-spaces $X \to Map_{\mathrm{Sp}}(E,E')$. This statement cannot be true as is, because $Map_{\mathrm{Sp}}(E,E')$ has no natural $E_\infty$-structure. Instead, it has an $\infty$-operad structure. More precisely, there exists an $\infty$-opead $\mathcal{O}_{E,E'}$ whose colors are the spectra maps $E \to E'$, and such that for $f_1,...,f_n \in Map_{\mathrm{Sp}}(E,E')$ and a target $f_0 \in Map_{\mathrm{Sp}}(E,E')$ the space of multi-maps $(f_1,...,f_n) \to f_0$ is the space of homotopies from the composition $E^{\wedge n} \stackrel{f_1 \wedge ... \wedge f_n}{\longrightarrow} (E')^{\wedge n} \to E'$ to the composition $E^{\wedge n} \to E \stackrel{f_0}{\to} E'$. Note that $E_\infty$-algebras in $\mathcal{O}_{E,E'}$ are exactly the $E_\infty$-maps $E \to E'$. Since $X$ is an $E_\infty$-space we can also think of it as an $\infty$-operad $\mathcal{O}_X$ whose colors are the points of $X$ and such that for $x_1,...,x_n \in X$ and a target point $x_0 \in X$ the space of multimaps $(x_1,...,x_n) \to x$ is the space of paths from $x_1\cdot... \cdot x_n$ to $x_0$. One can then show that
$$ Map_{Alg_{E_\infty}(\mathcal{C})}(E_X,E'_X) \simeq Map_{\infty-operads}(\mathcal{O}_X,\mathcal{O}_{E,E'}) $$
It so happens that in this particular case this space of operad maps breaks into a product $Map_{E_\infty}(X,Map_{\mathrm{Sp}}(S^0,E')) \times Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(E,E')$ as described above. To understand why this happens, we note that the $\infty$-operad $\mathcal{O}_X$ has a unit, i.e., it has a color which corepresents the space of $0$-ary maps, namely, the color corresponding to the unit $1 \in X$. This color carries a canonical structure of an $E_\infty$-object in $\mathcal{O}_X$. In particular, every $\infty$-operad map $\mathcal{O}_X \to \mathcal{O}_{E,E'}$ will send $1$ to an $E_\infty$-algebra object in $\mathcal{O}_{E,E'}$, i.e., to a map of $E_\infty$-algebras $f:E \to E'$. We hence obtain a map of spaces
$$ p:Map_{\infty-operads}(\mathcal{O}_X,\mathcal{O}_{E,E'}) \to Map_{Alg_{E_\infty}(\mathrm{Sp})}(E,E') $$
Now for every $E_\infty$-algebra object $f$ of $\mathcal{O}_{E,E'}$ one can form the $\infty$-operad $\mathrm{Mod}_f(\mathcal{O}_{E,E'})$ of $f$-modules. One can then show that the homotopy fiber of $p$ over $f$ is equivalent to the subspace of $\infty$-operad maps $\mathcal{O}_X \to \mathrm{Mod}_f(\mathcal{O}_{E,E'})$ which send $1$ to the $f$ (considered as an $f$-module). Now happens something which is very specific to this situation: the $\infty$-operads $\mathrm{Mod}_f(\mathcal{O}_{E,E'})$ are canonically equivalent to each other for all $f$! The reason is that, just like $E_\infty$-algebra objects in $\mathcal{O}_{E,E'}$ correspond to $E_\infty$-algebra maps $E \to E'$, modules over $f$ correspond to "affine maps" $E \to E'$ whose "linear part" is $f$, i.e., maps $g: E \to E'$ which loosely speaking satisfy $g(xy) = f(x)g(y)$. Such maps are all given by the informal formula $x \mapsto f(x)\cdot b$. More precisely, evaluation at the unit $S^0 \to E$ induces an equivalence between the space of $f$-modules and the space of maps $S^0 \to E'$. We then see that for every such $f$, the $\infty$-operad $\mathrm{Mod}_f(\mathcal{O}_{E,E'})$ is canonically equivalent to the $\infty$-operad associated with the $E_\infty$-space $Map_{\mathrm{Sp}}(S^0,E')$. We then obtain a canonical splitting
$$ Map_{Alg_{E_\infty}(\mathcal{C})}(E_X,E'_X) \simeq Map_{E_\infty}(X,Map_{\mathrm{Sp}}(S^0,E')) \times Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(E,E') $$
