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To address Hanno's question about checking that composition gives a graded-commutative ring structure on $End^{*}(\mathbb{Z}) = \oplus_i [\mathbb{Z}, \Omega^{-i} \mathbb{Z}]$ suppose first that
$a \stackrel{f}{\to} b \stackrel{g}{\to} c \stackrel{h}{\to} \Omega^{-1} a$
is a distinguished triangle in the stable category. Then we can produce from this two isomorphic triangles: one by rotation namely
$a \stackrel{-\Omega^{-1}f}{\to} b \stackrel{-\Omega^{-1}g}{\to} c \stackrel{-\Omega^{-1}h}{\to} \Omega^{-1} a$
and one by applying $\Omega^{-1}\mathbb{Z} \otimes_\mathbb{Z}$
$\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} a \stackrel{1_{\Omega^{-1}\mathbb{Z}}\otimes f}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} b \stackrel{\Omega^{-1}\mathbb{Z}\otimes g}{\to}\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} c \stackrel{\Omega^{-1}\mathbb{Z}\otimes h}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} \Omega^{-1}a$
The point of this is that the natural isomorphism commuting the loops functor across introduces a sign change, which is precisely the one you pick up by changing the order of composition since changing the composition order is equivalent to applying symmetry isomorphisms to the tensor product which is equivalent to commuting loops across.

To be completely explicit about this there are two functors naturally isomorphic to $\Omega^{-1}$ namely $\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z}(-)$ and $\mathbb{Z}\otimes \Omega^{-1}(-)$ since $\otimes_\mathbb{Z}$ is biexact there are natural transformations commuting $\Omega^{-1}$ with $\otimes_\mathbb{Z}$ namely
$\Omega^{-1}(-)\otimes_\mathbb{Z} (-) \stackrel{\sim}{\rightarrow} \Omega^{-1}((-)\otimes_\mathbb{Z}(-)) \stackrel{\sim}{\leftarrow} (-)\otimes_\mathbb{Z} \Omega^{-1}(-)$
Our example triangles above which can be obtained from one another by first moving the loops to the right and then applying the unit transformation show that there must be a sign attached to this map for this to give an isomorphism of these triangles. In particular for $\otimes_\mathbb{Z}$ to be compatible with the triangulated structure the two natural isomorphisms moving $\Omega$ about must have different signs. These natural isomorphisms are the precise cause of the sign change.

1

To address Hanno's question about checking that composition gives a graded-commutative ring structure on $End^{*}(\mathbb{Z}) = \oplus_i [\mathbb{Z}, \Omega^{-i} \mathbb{Z}]$ suppose first that
$a \stackrel{f}{\to} b \stackrel{g}{\to} c \stackrel{h}{\to} \Omega^{-1} a$
is a distinguished triangle in the stable category. Then we can produce from this two isomorphic triangles: one by rotation namely
$a \stackrel{-\Omega^{-1}f}{\to} b \stackrel{-\Omega^{-1}g}{\to} c \stackrel{-\Omega^{-1}h}{\to} \Omega^{-1} a$
and one by applying $\Omega^{-1}\mathbb{Z} \otimes_\mathbb{Z}$
$\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} a \stackrel{1_{\Omega^{-1}\mathbb{Z}}\otimes f}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} b \stackrel{\Omega^{-1}\mathbb{Z}\otimes g}{\to}\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} c \stackrel{\Omega^{-1}\mathbb{Z}\otimes h}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} \Omega^{-1}a$
The point of this is that the natural isomorphism commuting the loops functor across introduces a sign change, which is precisely the one you pick up by changing the order of composition since changing the composition order is equivalent to applying symmetry isomorphisms to the tensor product which is equivalent to commuting loops across.