Let $F$ be a multiplicative generalized cohomology theory ($\tilde F$ will denote the reduced theory as usual).
Let $\sigma^0 \in F^0(pt) \simeq \tilde{F}^0(S^0)$, and let for any $n\in \mathbb{Z}$, $\sigma^n \in \tilde{F}^n(S^n)$ be the element associated to $\sigma^0$ under the suspension isomorphism.
In the famous paper of Atiyah, Bott and Shapiro (Clifford Modules) paragraph 12, it is said that $$ \tilde{F}^\bullet (S^n) = \oplus_q \tilde F^q(S^n)$$ is a free module over $\tilde{F}^\bullet (S^0)$ generated by $\sigma^n$.
Why is that true? It is not clear to me what is the relation between the suspension isomorphism and the cup product induced by the multiplication.
Naively we would like to show that $$S^n(x \cup \sigma^0) = S^n(x) \cup S^n(\sigma^0)= S^n(x) \cup \sigma^n,$$ where $S^n$ is the suspension isomorphism and $x \in \tilde F^k(S^0)$.
If true, how does the above equality follow from the axioms defining the multiplication? I am taking as my definition that of https://ncatlab.org/nlab/show/multiplicative+cohomology+theory