Norm on unitisation of a $C^\ast$ algebra In the theory of $C^\ast$ algebras there exists the following theorem:
If $A$ is a $C^\ast$ algebra and $\widetilde{A}$ denotes its unitisation then there exists exactly one norm that extends the norm on $A$ and makes $\widetilde{A}$ into a $C^\ast$ algebra. 
This theorem is proved for example in Murphy on page 40. The only property of a $C^\ast$ algebra that is not already satisfied by the usual norm on $\widetilde{A}$ is $\|a\|^2 = \|a^\ast a\|$. The usual norm on $\widetilde{A}$ is defined as $\|(a,\lambda)\| = \|a\| + \lambda$. Let's for the sake of this question assume that $A$ is unital with unit $e$. Then the proof in Murphy endows $\widetilde{A}$ with the norm $\|(a,\lambda)\| = \|a + \lambda e \| + |\lambda|$. 
Then we should have: 
$$ \|(a,\lambda)\|^2 = \|(a,\lambda)^\ast (a,\lambda)\|$$
I tried to verify this but it seems to be wrong:
$$ \|(a,\lambda)\|^2 = (\|a + \lambda e\| + |\lambda|)^2 = \|a + \lambda e\|^2 + 2 |\lambda|  \|a + \lambda e\|+ |\lambda|^2$$ 
but
$$\begin{align} 
\|(a,\lambda)^\ast (a,\lambda)\| &= \|(a^\ast a + \lambda a^\ast + \overline{\lambda}a, \lambda \overline{\lambda})\| \\
&= \| a^\ast a + \lambda a^\ast + \overline{\lambda}a + \lambda \overline{\lambda} e\| + |\lambda|^2 \\
&=\|a^\ast (a + \lambda) + \overline{\lambda}(a + \lambda)\| + |\lambda|^2\\
&=\|(a^\ast + \overline{\lambda})(a + \lambda)\| + |\lambda|^2 \\
&= \|(a+\lambda)^\ast (a + \lambda)\| + |\lambda|^2\\
&=  \|(a+\lambda)\|^2 + |\lambda|^2\\
&\neq \|a + \lambda e\|^2 + 2 |\lambda|  \|a + \lambda e\|+ |\lambda|^2
\end{align}$$
What am I doing wrong here?

The norm on the direct sum $A \oplus B$ of two C*-algebras is the maximum norm $$\lVert (a,b) \rVert = \max\{\lVert a\rVert, \lVert b\rVert\},$$ not the $\ell_1$-norm that you write: an example that helps me remember this is to consider $A=C(K)$, $B = C(L)$ and to note that $A \oplus B$ should be isomorphic to $C(K \sqcup L)$.
Since Murphy identifies $\widetilde{A}$ with $A \oplus \mathbb C$ via $(a,\lambda) \mapsto (a + \lambda, \lambda)$, the norm you should consider on
$\widetilde{A}$ is $\lVert(a,\lambda)\rVert = \max\{\lVert a+\lambda\rVert,\lvert\lambda\rvert\}$ and a small modification of your calculations shows that this works.

Since Murphy doesn't spell it out explicitly: In the non-unital case, the norm on the unitisation is
$$
\lVert (a,\lambda) \rVert = \sup_{\lVert b \rVert \leq 1} \lVert ab+\lambda b\rVert.
$$
