Let $A$ be a unital $C^*$-algebra. I try to show that the map $$\psi: A \to A \otimes \mathbb{C}1_A: a \mapsto a \otimes 1_A$$ is a $*$-isomorphism. Here the tensor product of $C^*$-algebras is uniquely determined since $\mathbb{C}1_A$ is abelian (hence nuclear).
Consider the $*$-morphisms $$\pi_1: A \to A: a \mapsto a, \quad \pi_2: \mathbb{C} 1_A \to A: a \mapsto a$$ Then every element in $\pi_1(A)$ commutes with every element in $\pi_2(A)$, so the universal property of the maximal $C^*$-tensor product gives a unique $*$-morphism $$\pi: A \otimes \mathbb{C}1_A \to A$$ such that $\pi(a \otimes \lambda 1_A) = \pi_1(a) \pi_2(\lambda 1_A) = a \lambda 1_A = \lambda a$. Clearly $$\psi \circ \pi = 1 = \pi \circ \psi$$ and we are done, as this shows that $\psi$ is bijective.
However, the use of nuclearity and these results seems a bit overkill. Is this proof correct and is there an easier way?
There's definitely no need to appeal to nuclearity; and yes, the argument is insane overkill. The key observation that one needs is that $A\otimes\mathbb C\,1_A=A\otimes 1_A$ (that is, no sums are needed to form the algebraic tensor product). This is because $$ \sum_j a_j\otimes \lambda_j\,1_A=\sum_j(\lambda_ja_j)\otimes 1_A=\Big(\sum_j \lambda_ja_j\Big)\otimes 1_A. $$ This gives the immediate consequence that the tensor product agrees with the algebraic tensor product. Now you can define, as you did, $\psi:A\to A\otimes 1_A$ by $\psi(a)=a\otimes 1_A$. This is trivially verified to be a $*$-homomorphism and a bijection, so it is a C$^*$-isomorphism.