The answer is *no*, in general. Here is a counterexample:

Let $A$ be the algebra of bounded linear operators on $\ell^2(\mathbb{N})$, and let $a \in A$ be the left shift on $\ell^2(\mathbb{N})$. Then the spectrum of $a$ is the closed unit disk $\overline{D}$, and the point spectrum of the operator $a$ is the open unit disk $D$.

Now we note that the notion "point spectrum" - which is defined for operators - can be translated into a notion that makes sense in Banach algebras: For each $\lambda$ in the point spectrum $D$ of $a$ there exists an element $a_\lambda \in A \setminus \{0\}$ such that $(\lambda - a) a_\lambda = 0$. Indeed, let $x_\lambda \in \ell^2(\mathbb{N})$ be an eigenvector of the operator $a$ and choose an arbitrary non-zero functional $\varphi_\lambda$ on $\ell^2(\mathbb{N})$. Then the operator $a_\lambda := \varphi_\lambda \otimes x_\lambda \in A$, given by
$$
a_\lambda x = \langle \varphi_\lambda, x \rangle x_\lambda \qquad \text{for } x \in \ell^2(\mathbb{N}),
$$
satisfies $(\lambda - a)a_\lambda = 0$.

Hence, if $B$ is a unital Banach algebra that contains $A$ as a subalgebra, we still have $(\lambda - a)a_\lambda = 0$ in $B$, so $\lambda - a$ cannot be invertible in $B$ for any $\lambda \in D$.