Happy to find the present posts to get introduced to a unified unitisation procedure!

I'd like to elaborate further$-$thus permitting myself a higher verbosity level$-$on Johannes Hahn's answer, based on UwF's proposal (in a comment to the OP).

Let as before $V=A\oplus\mathbb{C}$ normed by $\max\{\|\cdot\|_A,|\cdot|\}$, then the $C^*$-norm on $A_1$ reads
$$
\sup\big\{\max\{\|ab+\lambda b\|_A,|\lambda z|\}\;\big|\; \max\{\|b\|_A, |z|\} = 1 \big\}
$$
$b$ and $z$, satisfying $\|b\|_A=1$ and $|z|=1$, may be varied independently to obtain
$$
=\;\sup\big\{\max\{\|ab+\lambda b\|_A,|\lambda|\}\;\big|\; \|b\|_A = 1\big\}\; =
$$

$$
\max\Big\{|\lambda|\:,\:\sup\big\{\|ab+\lambda b\|_A\,\big|\,\|b\|_A = 1\big\}\Big\}\; = \;\|(a,\lambda)\|_{A_1}
$$
This expression covers both the cases "$A$ is unital" and "$A$ has no unit".
The second argument to '$\max$' equals the operator norm of $L_a + \lambda\,\text{id}_A\in\mathscr{L}(A)$, where $L_a$ denotes left multiplication by $a$.

An important step consists in verifying $\forall x = (a,\lambda)\in A_1$ the $C^*$-condition $\|x\|^2_{A_1} = \|x^*x\|_{A_1}$ which is straightforward as declared in Johannes Hahn's answer:
For any $b$ with $\|b\|_A=1$ one has
$$
\|xb\|^2_A \le \|x^*xb\|_A \le \|L_{x^*x}\|_{\mathscr{L}(A)}
$$
hence the inequality
$$
\|x\|^2_{A_1} = \max\big\{|\lambda|^2,\|L_x\|^2_{\mathscr{L}(A)}\big\}
\le \max\big\{|\lambda|^2,\|L_{x^*x}\|_{\mathscr{L}(A)}\big\} = \|x^*x\|_{A_1}
$$
Because the operator norm is submultiplicative, one deduces
$\|x\|_{A_1}=\|x^*\|_{A_1}$ which in turn proves the $C^*$-condition.

**Remark:** Adopting the unitisation product as the action of $A_1$ on $V$, instead of the proposed one, yields failure in the case $A =\mathbb{C}$ already:

Consider the element $(-2,1)\in A_1$. Its square is $(0,1)$, whence
$$
\|(-2,1)^*(-2,1)\|_{A_1} = 1
$$
but
$$
\|(-2,1)\|_{A_1}\; =\; \sup\big\{\|(-2,1)(a,\lambda)\| \big| \max\{|a|,|\lambda|\}=1\big\}
$$
$$
= \;\sup\big\{\|(-a-2\lambda,\lambda)\| \big| \max\{|a|,|\lambda|\}=1\big\}
\; =\; 3
$$

$\quad\Longrightarrow\; C^*$-*condition is not fulfilled*.

**A kind of Archive**

The following case distinction stems from the primary version of this answer;
as it may still be helpful it is kept here:

**$\quad A$ has no unit**

Then the second value entering the '$\max$' in the gray block above *equals* the desired $C^*$-norm
$$\|(a,\lambda)\|_{A_1}
= \sup\big\{\|ab+\lambda b\|_A\,\big|\,\|b\|_A = 1\big\}$$

see e.g. [Pedersen: "Analysis now", 4.3.9 Lemma], also for proving the $C^*$-condition.

Since $A_1\twoheadrightarrow\mathbb{C}, (a,\lambda)\mapsto\lambda$ is a *-homomorphism, thus norm-nonincreasing, one has $ |\lambda| \le \|(a,\lambda)\|_{A_1}$. Hence considering the maximum is redundant.

**$\quad A$ is unital**

with unit $e$, then its unitisation $A_1$ is (isomorphic to) a direct sum of $C^*$-algebras
$$A_1\overset{\cong}{\longrightarrow} A\oplus\mathbb{C},\; a+\lambda 1\longmapsto (a+\lambda e,\lambda(1-e)),$$

cf. [Murphy: "C*-Algebras and Operator Theory", 2.1.6 Thm], and
$$\|(a,\lambda)\|_{A_1} = \max\big\{ \|a+\lambda e\|_A , |\lambda|\big\}$$
does the job as $C^*$-norm on $A_1$ (this rectifies a statement in the OP).

It coincides with the general expression above because (left) regular representations of $C^*$-algebras are isometric, i.e.,
$\|L_a\|_{\mathscr{L}(A)} = \|a\|_A$ holds $\forall a$ in any $A$

(to show equality: "$\le$" only depends on the norm being submultiplicative,

whereas "$\ge$" results from choosing $a^*/\|a\|$ if $a\ne 0$, combined with the $C^*$-condition).

Making the operator norm definition explicit within the current situation yields
$$\|a+\lambda e\|_A = \sup\big\{\|(a+\lambda e)b\|_A\,\big|\,\|b\|_A = 1\big\}$$

alwaysbe bigger than $A$. Exactly because the unital/non-unital distinction on the input is completely unnatural when one wants to define a functor $\{C^\ast-algebras\}\to\{C^\ast-algebras with one\}$. Also the relation $(C_0(X))_1 = C(X^+)$ demands that, because the one-point-compactification adds a point whether or not $X$ is compact. I'll try to have look in Murphys book as soon as possible. $\endgroup$ – Johannes Hahn Jun 24 '15 at 16:27notsaying that such a construction is better than the classical proof, for this reason I don't want to turn it into an answer, either. $\endgroup$ – UwF Jun 26 '15 at 6:5514more comments