The maximal tensor product satisfies the following universal property:
Let $A, B,$ and $C$ be C$^*$-algebras. If $\phi: A \rightarrow C$ and $\psi: B \rightarrow C$ are $*$-homomorphisms whose images commute, then there's a unique $*$-homomorphism $\phi \otimes \psi : A\otimes_{max} B \rightarrow C$ such that $(\phi\otimes \psi) (a \otimes b) = \phi(a)\psi(b)$. See page 193 of Murphy's book for a proof here.
If we take $A \otimes_{max} B$ with inclusions $i_A (a) = a \otimes 1$ and $i_B(b) = 1 \otimes b$, then the universal property above guarantees that we have found our coproduct in C$^*$-alg$_{com}^1$. Note that the "max" in the previous sentence wasn't necessary, since $A$ and $B$ are nuclear anyway. So you could just as easily take the spatial tensor product, if you like that construction better.