The setup is as follows. $A$ is a commutative Banach algebra, and $\Delta_A$ is the structure space of $A$, consisting of all non-zero continuous algebra homomorphisms $f:A\to \Bbb C$. The text says that
For a given multiplicative functional $m \in \Delta_A$, there exists precisely one extension of $m$ to a multiplicative functional on $A^e$ given by $$m^e(a,\lambda) = m(a) + \lambda$$
where $A^e = A\times \Bbb C$ is the unitization of $A$, with multiplication defined as $$(a,p)\cdot(b,q) = (ab + pb + qa, pq)$$ $A^e$ is also a Banach algebra.
Question: Why does a multiplicative functional $m \in \Delta_A$ extend uniquely to $m^e \in \Delta_{A^e}$?
Note that $A$ embeds into its unitization as $A \cong A\times \{0\} \hookrightarrow A \times \Bbb C = A^e$. For any extension $\widetilde m$ of $m$, we require $\widetilde m(a,0) = m(a)$ for all $a\in A$. Perhaps using the multiplicative structure of $A^e$, we must show that the extension $\widetilde m$ is unique. Then, since $m^e(a,\lambda) = m(a) + \lambda$ defined above is indeed in $\Delta_{A^e}$ and restricts to $m$ on $A$, we are done. How should I go about this?
Reference: Pg. $45$, Principles of Harmonic Analysis by Anton Deitmar and Siegfried Echterhoff.
One can explicitly show that $m^e$ must be of the required form. Suppose $m^e$ is an extension of $m:A\to\Bbb C$ to $A^e$. Then, $$(a, \lambda) = (a, 0) + (0, \lambda) = (a,0) + \lambda (0,1)$$ and $$m^e(a, \lambda) = m^e(a,0) + \lambda m^e(0,1) = m(a) + \lambda$$ since $m^e(0,1) = 1$ (recall that $(0,1)$ is the identity in $A^e$) and $A \cong A \times \{0\}$ gives the identification $a = (a,0)$.