I think it goes like this (I am only dealing with the unital case, I am not exactly sure how this works in the non-unital case).

Let $(A_{i})_{i\in I}$ be an inductive system of unital $C^{\ast}$-algebras with unital connecting homomorphisms $f_{ij}: A_{i} \to A_{j}$ and let's denote the limit by $A$. If $B$ is another unital $C^{\ast}$-algebra, we would like to show that the inductive limit of $(A_{i}\otimes_{\rm{max}} B)_{i\in I}$ is isomorphic to $A\otimes_{\rm{max}} B$. In order to do so, we will show that $A\otimes_{\rm{max}}B$ has the required universal property, i.e. any compatible family of $\ast$-homomorphisms from $A_{i}\otimes_{\rm{max}} B$ to a unital $C^{\ast}$-algebra $C$ gives rise to a $\ast$-homomorphism from $A\otimes_{\rm{max}} B$. Note now that, by the universal property of the maximal tensor product, a $\ast$-homomorphism $\varphi_{i}: A_{i}\otimes_{\rm{max}} B \to C$ is given by $\varphi_{i} = \theta_{i}\cdot \psi_{i}$, where $\theta_{i}:A_{i} \to C$ and $\psi_{i}: B \to C$ are $\ast$-homomorphisms with commuting ranges. As connecting maps are of the form $f_{ij}\otimes \rm{Id}$, we can check that $\psi_{i}=\psi_{j}$, which we call $\psi$ from now on, and $\theta_{i} = \theta_{j}\circ f_{ij}$. Since $A$ is the limit of $A_i$'s, we get a $\ast$-homomorphism $\theta:A\to C$. As the union of ranges of $A_i$'s inside $A$ is dense, this $\ast$-homomorphism has range commuting with the range of $\psi$, so we get a $\ast$-homomorphism $\varphi: A\otimes_{\rm{max}} B \to C$ given by $\varphi(x\otimes y):= \theta(x)\psi(y)$.