The following argument seems easier, but there might be a still more fundamental one.
Notice that $ \phi: A^{\sim} \to \mathbb{C} $ above is also a $ C^{*} $-algebraic homomorphism. As $ C^{*} $-algebraic homomorphisms are automatically contractive (which is a consequence of a not-too-difficult spectrum argument), we have
$$
\forall (a,\lambda) \in A \times \mathbb{C}: \quad
|\lambda|
= |\phi(a,\lambda)|
\leq \| (a,\lambda) \|_{A^{\sim}}
= \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| a b + \lambda \cdot b \|_{A}.
$$
No hard facts about positive linear functionals on $ C^{*} $-algebras were used.
Additional Information:
The inequality is valid if we only assume that $ A $ is a Banach algebra that satisfies the following conditions:
- $ A $ has no two-sided identity.
- $ A $ has a right approximate identity (r.a.i.) norm-bounded by $ 1 $, which we denote by the net $ (e_{i})_{i \in I} $.
Let $ A $ be a Banach algebra satisfying Conditions (1) and (2).
Let $ \mathbb{L}(A) $ denote the Banach algebra of all bounded homomorphisms from $ A $ to itself, where multiplication is defined by composition and the norm is simply the operator norm.
The only leap (not too great, I hope!) of imagination required is to notice that for $ (a,\lambda) \in A \times \mathbb{C} $,
$$
\sup_{b \in A, ~ \| b \|_{A} \leq 1} \| a b + \lambda \cdot b \|_{A}
$$
represents the operator norm of $ L_{a} + \lambda \cdot \text{id}_{A} \in \mathbb{L}(A) $, where $ L_{a} $ denotes left-multiplication by $ a $.
Define
- $ A^{\sim} \stackrel{\text{df}}{=} \{ L_{a} + \lambda \cdot \text{Id}_{A} \mid (a,\lambda) \in A \times \mathbb{C} \} $,
- $ L_{A} \stackrel{\text{df}}{=} \{ L_{a} \mid a \in A \} $.
Then clearly $ A^{\sim} $ is a sub-algebra of $ \mathbb{L}(A) $.
Claim 1: $ L_{A} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $.
Proof of Claim 1
Let $ a \in A $. We already know that $ \| L_{a} \| \leq \| a \|_{A} $. However, $ \| e_{i} \|_{A} \leq 1 $ for all $ i \in I $ and
$$
\lim_{i \in I} \| {L_{a}}(e_{i}) \|_{A}
= \lim_{i \in I} \| a e_{i} \|_{A}
= \| a \|_{A},
$$
so we actually have $ \| L_{a} \| = \| a \|_{A} $. The map $ x \mapsto L_{x} $ is therefore a bijective isometry from $ A $ to $ L_{A} $, so $ L_{A} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. $ \quad \blacksquare $
Define a (not a priori continuous) linear functional $ \phi $ on $ A^{\sim} $ by
$$
\phi(L_{a} + \lambda \cdot \text{Id}_{A}) \stackrel{\text{df}}{=} \lambda.
$$
To prove that $ \phi $ is well-defined, we must show that given $ (a,\lambda) \in A \times \mathbb{C} $, if $ L_{a} + \lambda \cdot \text{Id}_{A} = 0_{\mathbb{L}(A)} $, then it must follow that $ \lambda = 0 $.
Assume the contrary. If we define $ e \stackrel{\text{df}}{=} - \dfrac{1}{\lambda} \cdot a $, then $ L_{e} = \text{Id}_{A} $, which means that $ e $ is a left identity of $ A $. Now, for all $ x \in A $, we have
\begin{align}
x e - x
& = \lim_{i \in I} ~ (x e - x) e_{i} \qquad
(\text{As $ (e_{i})_{i \in I} $ is an r.a.i..}) \\
& = \lim_{i \in I} ~ (x e e_{i} - x e_{i}) \\
& = \lim_{i \in I} ~ (x e_{i} - x e_{i}) \qquad (\text{As $ e $ is a left identity.}) \\
& = \lim_{i \in I} ~ 0_{A} \\
& = 0_{A}.
\end{align}
Hence, $ e $ is a right identity of $ A $ as well. This contradicts our assumption that $ A $ has no two-sided identity, so we indeed have $ \lambda = 0 $.
Claim 2: $ A^{\sim} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $.
Proof of Claim 2
As $ L_{A} $ is a complete linear subspace of $ A^{\sim} $, it is automatically closed. The quotient vector space $ A^{\sim} / L_{A} $ can thus be given a norm, which then has to be complete because $ \ker(\phi) = L_{A} $ and so
$$
A^{\sim} / L_{A}
= A^{\sim} / \ker(\phi)
\cong \mathbb{C}.
$$
Exploiting the result that a normed vector space is complete if its quotient by a complete linear subspace is complete, we conclude that $ A^{\sim} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. $ \quad \blacksquare $
We now see that $ A^{\sim} $ is a unital Banach algebra w.r.t. the operator norm on $ \mathbb{L}(A) $. As $ \phi $ is a multiplicative linear functional on $ A^{\sim} $, it follows that $ \phi $ must be bounded with norm $ \leq 1 $ (this is an easy fact whose proof can be found in Rudin’s Real and Complex Analysis).
The inequality is therefore established.
Notes:
If Condition (1) is violated, i.e., $ A $ has a two-sided identity $ e $, then the inequality is false because then
$$
1
\nleq \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| e b - 1 \cdot b \|_{A}
= 0.
$$
I do not know what happens if Condition (2) is violated, however.
If $ A $ is a $ C^{*} $-algebra, then we only need Condition (1) because Condition (2) automatically holds. This can be misleading, however, because it is not at all obvious that non-unital $ C^{*} $-algebras should have an r.a.i. norm-bounded by $ 1 $. Fortunately, it turns out that in proving Claims 1 and 2, the $ C^{*} $-identity is all that is needed as it takes over the pivotal role played by Condition (2).
