In W.Rudin's book P295, Theorem 11.28 is about the positive element(i.e. $a\geq 0$), the property: $\mathscr{A}$ is $C^{*}$-algebra,
(e)If $a\in \mathscr{A}$, then $aa^{*}\geq 0$.
(f)If $a\in \mathscr{A}$, then $e+aa^{*}$ is invertible in $\mathscr{A}$.
Q1: I have other ways to approach the (e), but I do not totally sure it works. My method is in the same way of W.Rudin at the beginning, considering the Gelfand transform $\Gamma$ in a maximal normal set of $a$ $\mathscr{B}\subset \mathscr{A}$, we have
$$\Gamma(aa^{*})=\Gamma(a)\Gamma(a^{*})=\vert \Gamma(a)\vert^{2} \geq 0,$$ then $aa^{*} \geq 0.$
Q2: How to use the (e) to show (f)?
$e+aa^{*}$ is invertible if and only if $\Vert aa^{*}\Vert \leq 1$.How to use it to show the (f)?
For your first question, there isn't necessarily a maximal normal set containing $a$. The reason for this is because generally $a$ doesn't commute with $a^*$. Thus we have a bit more work to prove that $aa^*\geq0$ (hence the longer proof in Rudin).
For your second question, let $\mathscr B$ denote the C$^*$-subalgebra of $\mathscr A$ generated by $aa^*$ (or the maximal normal set containing $aa^*$, if you prefer). Note that if $e+aa^*$ is invertible in $\mathscr B$, then $e+aa^*$ is invertible in $\mathscr A$. Furthermore, $e+aa^*$ is invertible in $\mathscr B$ if and only if $\gamma(e+aa^*)\neq0$ for all complex homomorphisms $\gamma$ on $\mathscr B$ (for then $0\notin\sigma_\mathscr{B}(e+aa^*)$). But we have $$\gamma(e+aa^*)=\gamma(e)+\gamma(aa^*)\geq1+0>0$$ for all complex homomorphisms $\gamma$, and the result follows.