I have not yet considered the new answer in full detail, so apologies for not addressing it. I'm coming back to this question after a while, so I thought I'd share some observations which came up during that time, and which might interest someone (at least, those someones who thought the original question was of any interest).
Example (3) above is actually a particular case of Example (5) with $p=2$, in the sense that it is obtained by restricting from 2-pro-finite spaces to pro-finite sets. Indeed, recall the notion of "solvable" which appears in Lurie's $p$-adic homotopy theory paper. In the discrete case, the data of a solvable $\overline{\mathbb{F}}_2$-algebra is essentially equivalent to the data of a Boolean algebra. In fact, example (3) admits a version for every prime $p$, replacing Boolean algebras with algebras which satisfy the relation $a^p=a$ for every $a$. These categories of "$p$-Boolian algebras" are in fact all equivalent to the opposide category of pro-finite sets (or totally disconnected compact Hausdorff spaces).
In every context where there is an interesting notion of geometric objects and commutative algebra objects which are connected by some duality, there is usually also a related category of "abelian group objects" underlying the situation. In particular, the commutative algebras are commutative algebra objects in this category, and there is typically an adjunction between geomtric objects and abelian group objects which associates to an abelian group its "underlying space" and to a geometric obejct the free abelian group generated from it. This later abelian group should be dual in some since to the underlying abelian group of the algebra of functions on the space. The adjunction between geometric objects and abelian group objects then induces a comonad on the category of abelian group objects. In that case one can ask two types of questions: (i) to what extent is this adjunction comonadic? in other words, how close is it to exhibiting spaces as coalgebras over this comonad? and (ii) how close is this comonad to being the cocommutative coalgebras comonad? When both answers are sufficiently positive we get that spaces look a lot like cocommutative coalgebras in abelian groups. Since abelian groups often have something that is close to self duality, this makes spaces close to the opposite of commutative algebras.
If one is working in the $\infty$-categorical setting, and $\mathcal{C}$ is some $\infty$-category of geometric objects, then one has a natural choice for the $\infty$-category $\mathcal{A}$ of abelian group objects, namely, the stabilization $Sp(\mathcal{C})$ of $\mathcal{C}$. Alternatively, one can tensor the stabilization with a field, such as $\mathbb{Q}$ or $\overline{\mathbb{F}}$, the choices underlying examples (4) and (5) above. If $\mathcal{C}$ is furthermore presentable then we will have an adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$ where we can consider the left functor as an analogue of the free abelian group functor. If $\mathcal{C}$ carries a symmetric monoidal structure which preserves colimits in each variable separately then $\mathcal{A}$ will inherit the same type of structure and the left functor $\mathcal{C} \to \mathcal{A}$ will be monoidal.
Let $K: \mathcal{A} \to \mathcal{A}$ be the comonad corresponding to the adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$. Since $\mathcal{A}$ is monoidal it makes since to ask if $K$ is the coalgebra comonad of some operad enriched in $\mathcal{A}$. This question can be tackled using Goodwillie calculus and obstruction theory, see work of Gijs Heuts (e.g., https://arxiv.org/abs/1510.03304). In particular, these obstructions vanish when working over a field of characteristic $0$, so if we take $\mathcal{A}$ to be the stabilization tensored with $\mathbb{Q}$, then the comonad $K$ will always come from some operad in $\mathcal{A}$. The spaces of $n$-ary operations of this operad (considered as $\Sigma_n$-objects in $\mathcal{A}$) are given by the Goodwillie derivatives of $K$ of $\mathcal{A}$. For this operad to be the commutative operad we need that the derivatives are all the unit object. This is likely to happen for completely abstract reasons when the monoidal structure on $\mathcal{C}$ is the Cartesian one, i.e., when $\mathcal{C}$ is Cartesian closed (a typical property of categories of geometric objects). Here is a cool heuristic explanation of this phenomenon that I learned from Tomer Schlank: the right adjoint $\mathcal{A} \to \mathcal{C}$ preserves Cartesian products and under our assumptions the left adjoint sends Cartesian products to tensor products in $\mathcal{A}$. Since $\mathcal{A}$ is stable products and coproducts in $\mathcal{A}$ coincide, and are often known simply as direct sums. One may then conclude that the comonad $K: \mathcal{A} \to \mathcal{A}$ sends direct sums to tensor products. In other words, in the analogy between functors and functions underlying the Goodwille calculus, $K$ is analogous to a function $\mathbb{R} \to \mathbb{R}$ which sends addition to multiplication. In other words, $K$ is analogous to an exponential function. As such, its derivatives at $0$ are the sequence of powers of some number $a$. There is a natural family of operads which have the same property. If $A \in \mathcal{A}$ is a cocommutative bialgebra object then the theory of commutaive $A$-algebras is controlled by an operad whose $\Sigma_n$-object of $n$-ary operations is $A^{\otimes n}$. It seems likely (though I don't have a proof of this), then for every Cartesian closed $\infty$-category $\mathcal{C}$ there is a cocommutative bialgebra object in $\mathcal{A} = Sp(\mathcal{C}) \otimes \mathbb{Q}$ such that the comonad associated to the adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$ is the comonad of cocommutative $A$-coalgebras. For example, when $\mathcal{C}$ is the $\infty$-category of spaces this $A$ is the unit object of $\mathcal{A}$ and one gets the commutative operad.
Edit: A closer look at 1 shows that this $A$ is actually always the unit, see Lemma B.3. On the other hand, the situation is more complicated than I thought, so remark (4) should really be taken loosely.
