@DuchampGérardH.E.: Suppose some $(a,\lambda)\in A_1$ acts trivially. Then you apply it first to $(0,1)$ to see that $\lambda=0$ and then to $(a^*,0)$ to get $a=0$. I am not saying that such a construction is better than the classical proof, for this reason I don't want to turn it into an answer, either.

I thought of taking for $A\oplus\mathbb{C}$ the $C^*$-algebraic direct sum, i.e. $\|(b,\mu)\|=\max(\|b\|,|\mu|)$, and letting $A_1$ act on $A\oplus\mathbb{C}$ as $(a,\lambda)(b,\mu)=((a+\lambda)b,\lambda\mu)$.

Isnt't this more a question of cosmetics? We know what the $*$-algebra structure of $A_1$ should be, so there exists at most one norm that makes it a $C^*$-algebra. You could make the construction look more uniform, if you let it act on $A\oplus\mathbb{C}$.

@JohannesHahn: The proof in Murphy's book answers your question (and the one by the OP), I believe. There are several procedures available to make C*-algebras unital. Passing to the multiplier algebra is probably more elegant (and more functorial) - this is done in Theorem 2.1.5 in Murphy's book (but if you algebra did not have a unit, then it usually adds a lot more than just a unit). If you want to use $A_1$, then you will never keep your original algebra, even if it was already unital.

See Theorem 2.1.6 in the book by Murphy, he treats also the case where $A$ is already unital.