A bit of confusion on one thing: whenever I see an explanation for the existence of a graded-commutative multiplication on $\pi_*(R) = \bigoplus_n \pi_n(R)$ for a simplicial ring $R$, it's usually a sketchy explanation along the lines of: if $S^n \to R$ and $S^m \to R$ represent homotopy classes in $\pi_n(R)$ and $\pi_m(R)$, then $S^n \wedge S^m \to R \wedge R \to R$ (given by smashing the two maps together and then using multiplication in $R$) is the product, where $S^n$ and $S^m$ are supposed to be simplicial $n,m$-spheres.
But I don't understand the implicit definition here. I guess $S^1 = \Delta^1/\partial\Delta^1$. But then if you explicitly compute $S^1 \wedge S^1$, it seems to have a non-degenerate $1$-simplex $x$ and a degenerate one $*$, which don't appear to be homotopic (i.e. you can check that there aren't any $2$-simplices whose boundary is $(*,*,x)$). So this thing $S^1 \wedge S^1$ doesn't even appear to have the same homotopy type as $S^2 = \Delta^2/\partial\Delta^2$ (which is the thing you use to define homotopy classes in, e.g., Goerss-Jardine).
Furthermore, there also appear to be not one but two non-degenerate $2$-simplices in $S^1 \wedge S^1$. So how would a map $S^1 \wedge S^1 \to R$ give you a $2$-simplex?
I realized that this is addressed in Section 4.3 of "Notes on simplicial homotopy theory" by Joyal and Tierney, and that in fact you can construct weak equivalences $(S^1)^{\wedge n} \to \Delta^n/\partial\Delta^n$.