Let $A$ be a unital $C^\ast$ algebra and let $B$ be a (not necessarily unital) $C^\ast$-subalgebra such that $B \oplus \mathbb C = A$.
I want to argue that the map $\varphi : \widetilde{B} \to A$, $(b,\lambda) \mapsto b + \lambda$ is an isomorphism. It should be obvious but it's not obvious to me.
My thoughts: it is clearly a $\ast$-homomorphism hence I can show that it is isometric from which it follows that it is injective. But given $a\in A$ how can I construct $(b,\lambda)$ such that $b+\lambda =a$?
I also thought about using the isomorphism $B \oplus \mathbb C = A$. Then given $a \in A$, $a$ is identified with some $(b,\lambda)$. The problem is then that nothing guarantees that $b + \lambda = a$.
Maybe this will clarify things for you. Let $B=C_0(\mathbb R)$. Then you have, as a direct sum of C$^*$-algebras, $$ B\oplus\mathbb C=\left\{\begin{bmatrix}f&0\\ 0&\lambda\end{bmatrix}:\ f\in B,\ \lambda\in\mathbb C\right\}. $$ In that algebra the product is coordinate wise, and the algebra is not unital.
The algebra the Murphy uses in the pages you are using (44-45) is $B+\mathbb C\,1$, i.e. the unitization of $B$, $$ \{f+\lambda:\ f\in B, \ \lambda\in \mathbb C\}\simeq\{(f,\lambda):\ f\in B,\ \lambda\in\mathbb C\}, $$ the latter with the pointwise sum and the product $(f,\lambda)\cdot(g,\mu)=(fg+\lambda g+\mu f,\lambda\mu)$.