Plausibly we can construct examples from the power series expansion of formal group laws. We can try to write down such examples by writing down nice functions $q(x)$ and hoping that the formal group law

$$q^{-1} (q(x) + q(y)) = \sum s_{i, j} x^i y^j$$

has positive integer coefficients so it can be upgraded to an "analytic bifunctor"

$$\text{Set} \times \text{Set} \ni (A, B) \mapsto \bigsqcup_{i, j} S_{i, j} \times A^i \times B^j \in \text{Set}$$

which is a candidate to be a monoidal structure. For example,

- the disjoint union corresponds to choosing $q(x) = x$,
- the cartesian product corresponds to choosing $q(x) = \log x$,
- the third example corresponds to choosing $q(x) = \log (sx + 1)$.

The fourth example should also arise in this way with $q(x)$ related to the arctangent; the relevant formal group law is $\frac{x + y + 2s xy}{1 - s^2 xy}$ which resembles the tangent addition formula $\frac{x + y}{1 - xy}$. Speaking of which, this should give another family of examples (with $q(x) = \arctan sx$), namely

$$A \otimes B = (S^2 \times A \times B)^{\ast} \times (A + B).$$

**Edit:** Apparently we are already pretty close to exhausting all of the formal group laws that can be expressed as rational functions! See this blog post. It seems plausible, but I haven't checked, that

$$A \otimes B = (S^2 \times A \times B)^{\ast} \times (A + B + T \times A \times B)$$

works.

**Edit #2:** Hmm. I can only find an associator in the case already mentioned in the problem, where $T = S + S$. In this case, as mentioned in the link, $A \otimes B$ can be interpreted as the set of all words which start with an element of $A$ or $B$ and then alternate between $A$ and $B$, with the alternations delimited by elements of $S$, and ending with an element of $A$ or $B$. Then $(A \otimes B) \otimes C$ and $A \otimes (B \otimes C)$ both describe the corresponding construction where we don't repeat among the choices $A, B, C$ consecutively.

4more comments