An ordinary category $\mathcal{C}$.
For each small ordinal $\alpha$, a functor $T_\alpha : \mathcal{C}^\alpha \to \mathcal{C}$, which maps a $\alpha$-sequence $(A_0, A_1, \ldots)$ to $A_0 \otimes A_1 \otimes \cdots$.
A natural isomorphism $\textrm{id}_\mathcal{C} \Rightarrow T_1$.
For each partition $\alpha = \sum_{i < \gamma} \beta_i$, a natural isomorphism $$T_\gamma \circ (T_{\beta_0} \times T_{\beta_1} \times \cdots ) \Rightarrow T_\alpha$$ such that for each double partition $\alpha = \sum_{i < \gamma} \sum_{j < \delta_i} \beta_{i,j}$, where $\beta_i = \sum_{j < \delta_i} \beta_{i,j}$ and $\delta = \sum_{i < \gamma} \delta_i$, the two composites $$T_\gamma \circ (T_{\delta_0} \circ (T_{\beta_{0,0}} \times T_{\beta_{0,1}} \times \cdots) \times T_{\delta_1} \circ (T_{\beta_{1,0}} \times T_{\beta_{1,1}} \times \cdots) \times \cdots) \Rightarrow T_\gamma \circ (T_{\beta_0} \times T_{\beta_1} \times \cdots) \Rightarrow T_\alpha$$\begin{multline} T_\gamma \circ (T_{\delta_0} \circ (T_{\beta_{0,0}} \times T_{\beta_{0,1}} \times \cdots) \times T_{\delta_1} \circ (T_{\beta_{1,0}} \times T_{\beta_{1,1}} \times \cdots) \times \cdots) \newline \Rightarrow T_\gamma \circ (T_{\beta_0} \times T_{\beta_1} \times \cdots) \Rightarrow T_\alpha \end{multline} and \begin{multline} T_\gamma \circ (T_{\delta_0} \circ (T_{\beta_{0,0}} \times T_{\beta_{0,1}} \times \cdots) \times T_{\delta_1} \circ (T_{\beta_{1,0}} \times T_{\beta_{1,1}} \times \cdots) \times \cdots) \newline \Rightarrow T_\delta \circ ((T_{\beta_{0,0}} \times T_{\beta_{0,1}} \times \cdots) \times (T_{\beta_{1,0}} \times T_{\beta_{1,1}} \times \cdots) \times \cdots) \Rightarrow T_\alpha \end{multline} are equal, and the two composites $$T_\alpha \Rightarrow T_\alpha \circ (T_1 \times T_1 \times \cdots ) \Rightarrow T_\alpha$$ $$T_\alpha \Rightarrow T_1 \circ T_\alpha \Rightarrow T_\alpha$$ are the identity natural transformation on $T_\alpha$.
(Though, in some sense we assumed this in our definition...)
This can then be used to prove that the tensor product of modules makes the category of modules into an infinitary unbiased monoidal category.
Remark. Suppose we define infinitary monoids to be discrete infinitary strict monoidal categories. There are non-trivial small infinitary monoids: for example, take any complete small semilattice with $\sup$ as the monoid operation. However, any infinitary monoid that is also a group must be the trivial monoid: after all, for any element $x$, $$x^{-1} \cdot (x \cdot x \cdot x \cdot \cdots) = (x^{-1} \cdot x) \cdot (x \cdot x \cdot x \cdot \cdots) = x \cdot x \cdot x \cdot \cdots$$ and then cancel $x \cdot x \cdot x \cdot \cdots$ on both sides of the equation to obtain $x^{-1} = \textrm{id}$.
(EDIT) Non-example. The tensor product of modules doesn't give an infinitary monoidal product. To be precise, if we define $$\textrm{Hom}_R (A_0 \otimes_R A_1 \otimes_R A_2 \otimes_R \cdots, B) \cong \textrm{Multi}_R(A_0, A_1, A_2, \ldots ; B)$$ where by $R$-multilinear we mean that $f(\ldots, r a, \ldots) = r f (\ldots, a, \ldots)$, then I don't see how $R \otimes_R R \otimes_R R \otimes_R \cdots$ can be isomorphic to $R$. (A multilinear map $R \times R \times R \times \cdots \to B$ is not necessarily determined by just what it does to $(1, 1, 1, \ldots)$, unlike the finite case.)