Consider the Gelfand transform of $a$:
\begin{align*}\hat{a}:\Omega (A)&\to \sigma_A(a) \\
h &\mapsto \hat{a}(h)=h(a)
\end{align*}
The map is well defined and surjective, because for any element of a commutative unital Banach algebra we have
$$ \lambda \in \sigma_A(a)\iff \exists\;h\in \Omega(A):h(a-\lambda e)=0\iff \exists\;h\in \sigma_A(a):h(a)=\lambda$$
Moreover, since $A$ is generated by $a$, the map is also injective: indeed, since any $h\in \Omega(A)$ is a continuous algebraic homomorphism, the value of $h$ on $a$ uniquely determines $h$ on the closed subalgebra spanned by the element $a$, which by assumption is exactly $A$.
Finally, to show that it is a homeomorphism we only need to prove continuity, since $\Omega(A)$ is compact and $\sigma_A(a)$ is Hausdorff. But this is easy because $\Omega(A)$ has the weak*-topology, so if $\left\langle h_i\right\rangle_{i\in I}$ is a net that converges to $h$ in $\Omega(A)$ then $h_i(a)\to h(a)$ for all $a\in A$. (EDIT: This is also true on any commutative unital Banach algebra. It is the fact that the Gelfand transform of $a$ is also a bijection that allows us to conclude that it is in fact a homeomorphism).
