Let $A$ be a Banach algebra, and $A^e = A \times \Bbb C$ be its unitization, with multiplication defined as $$(a,p)\cdot(b,q) = (ab + pb + qa, pq)$$ $A^e$ is also a Banach algebra with $(0,1)$ as its unit. For unital Banach algebras, we define the spectrum $\sigma_A(a)$ of an element $a\in A$ as $$\sigma_A(a) = \{\lambda\in \Bbb C: \lambda\cdot 1_A - a \text{ is not invertible}\}$$ However, for Banach algebras without a unit, this is not possible. Consequently, we consider the unitization of $A$, and define $$\sigma_A(a) := \sigma_{A^e}(a)$$ in this case.
Question: If $A$ is a unital Banach algebra, we know that the unitization $A^e$ is isomorphic to $A \oplus \Bbb C$ via the map $(a,\lambda) \mapsto (a + \lambda \cdot 1_A, \lambda)$. In this case, how is $\sigma_A(a)$ related to $\sigma_{A^e}(a)$ for any $a\in A$? I suspect there must be some relationship between the two.