Why is $E^*(X)$ graded commutative? Context: I'm reading The stable homotopy category by Malkiewich (see his webpage, at Expository writings), and at this point the author hasn't actually introduced spectra, just the properties we would like the homotopy category of spectra to have. 
Based on this, he defined $E^*(X)$ for a CW-complex $X$ and a ring spectrum $E$ (only defined as a monoid object in the homotopy category for now); and an exercise is to prove that if $E$ is commutative (as a monoid object), then $E^*(X)$ (which we saw earlier had a graded ring structure) is graded commutative. 
I'm having trouble seeing how we can deduce that from the properties that were listed before, so I need some help for that. 
If I'm not mistaken, the ring structure on $E^*(X)= [\Sigma^\infty X,E]_{-*}$ can be defined as follows: given $f: \Sigma^n \mathbb S \to F(\Sigma^\infty X, E)$ and $g: \Sigma^m\mathbb S\to F(\Sigma^\infty X,E)$, we get $$\Sigma^{n+m}\mathbb S \to \Sigma^n \mathbb S \wedge \Sigma^m\mathbb S \to F(\Sigma^\infty X,E)\wedge F(\Sigma^\infty X,E) \to F(\Sigma^\infty X, E\wedge E) \to F(\Sigma^\infty X, E)$$
where the first map is given by $\Sigma^{n+m}\mathbb S = (\Sigma^\infty S^1)^{\wedge (n+m)}\wedge \mathbb S\to (\Sigma^\infty S^1)^{\wedge n}\wedge \mathbb S \wedge (\Sigma^\infty S^1)^{\wedge m}\wedge \mathbb S= \Sigma^n\mathbb S\wedge \Sigma^m\mathbb S$ (where I used the diagonal $\mathbb S\to \mathbb S\wedge \mathbb S$), the second map is given by the smash product of $f,g$, the third one is the obvious morphism, and the last one is pushforward through multiplication of $E$. 
Now if $E$ is commutative, the last arrows "don't matter" to get graded-commutativity, as they will only give commutativity, without sign. So to get the signs to come out right, I would need to work with the maps $\Sigma^{n+m}\mathbb S \to \Sigma^n \mathbb S \wedge \Sigma^m\mathbb S \to F(\Sigma^\infty X, E)\wedge F(\Sigma^\infty X, E)$
So I need to show (I think) that given maps $f: \Sigma^n \mathbb S\to Y, \Sigma^m\mathbb S\to Y$, the two maps I get $\Sigma^{n+m}\mathbb S\to Y\wedge Y$ are equal up to a sign $(-1)^{nm}$
Now  I wonder how general that last bit is : I know that when we work with topological spaces we have similar phenomena when we smash maps $S^n\wedge S^m$ and they're somehow due to some "reversal of orientation" (because everything is of the form $\Sigma X$ for some $X$, and we have a "reversal isomorphism" $\Sigma\to \Sigma$), but can it be seen "more fundamentally" at some level on, say, monoidal categories ? 
It would be so that I don't have to bother with the specifics of the situation that I don't understand so well (given that spectra haven't been defined at this point, and $\Sigma$ has been defined as $(\Sigma^\infty S^1)\wedge -$, but we haven't been given a lot of information about $\Sigma^\infty$ either)

So my question is first of all, am I correct about the definition of the ring structure; and then "how can we prove that it is graded commutative ?" but the best answer would be a general theorem about additive monoidal categories with a suspension functor or something along those lines, that tells us that whatever happens this is always graded commutative. 

If there's no such theorem, then something situation-specific is fine as well. 
 A: The point is indeed that the swap map $s:S^m\wedge S^n\to S^n\wedge S^m$ induces the natural automorphism of multiplication by $(-1)^{mn}$ on the corepresented functor $\pi_{m+n}$. This is because $(-1)^{mn}$ is the determinant of the map $\mathbb{R}^{m+n}\to \mathbb{R}^{m+n}$ swapping the last $n$ coordinates with the first $m$, and $s$ is the extension of this map to one-point compactifications. It's well known that linear maps with the same sign of their determinant are homotopic, so in short $s$ multiplies by $-1$ in $\pi_{m+n}S^{m+n}$, which gives my claim. To port this fact over to spectra, one simply applies the adjunction between $\Sigma^\infty$ and $\Omega^\infty$. 
Given that this apparently depends on rather special facts about the topology of the general linear group over the real numbers, it's a very interesting question whether there's an "abstract nonsense" argument. I do not think there is one in the generality of monoidal category theory, but in terms of $\infty$-category theory there is one, since there the construction of suspension has a universal property given by $SX=*\coprod_X *$, where the pushout is a homotopy pushout. Then any object which is a suspension is a cogroup in the homotopy category, and negation on the corepresented group-valued functor is induced by reversing the positions of the zero objects in the defining homotopy pushout. Then one can, with some care, define the coordinate-switching isomorphism abstractly, and show it coincides with negation. To see some arguments like this fleshed out, a good reference in Moritz Groth's paper on pointed and stable derivators. 
However, even in this broader context it's not necessary to make the more abstract argument-suspension can also be defined using the universal property of the $\infty$-category of (say, pointed) spaces, and then the argument in the first paragraph applies everywhere. 
