One way to prove the result is to first realise the $C^*$-algebra as sub-algebra of $\mathcal{B}(H)$ for a Hilbert space $H$ (GNS-construction) and then use the Lemmata below.
Lemma 1 Suppose $H$ is a Hilbert space and $S,T\in \mathcal{B}(H)$ are invertible, self-adjoint operators. Then $\Vert Sx\Vert \le \Vert T x \Vert$ for all $x\in H$ implies $\Vert T^{-1}x\Vert\le \Vert S^{-1} x\Vert$ for all $x\in H$.
Proof. Let $x,y\in H$, then
$$
\vert \langle T^{-1}x,y\rangle \vert = \vert \langle S^{-1}x,ST^{-1}y\rangle\vert\le \Vert S^{-1}x\Vert \cdot \Vert ST^{-1}y\Vert \le \Vert S^{-1}x\Vert \cdot\Vert y \Vert.
$$
Here we first used the Cauchy-Schwarz inequality and then the assumed estimate $\Vert Sz\Vert \le \Vert Tz \Vert$ for $z=T^{-1}y.$ The Lemma follows from taking the supremum over all $y\in H$ with $\Vert y \Vert = 1$.
Lemma 2 Suppose $H$ is a Hilbert space and $A,B\in \mathcal{B}(H)$ are invertible, self-adjoint operators. Then $A\le B$ implies $B^{-1}\le A^{-1}$.
Proof. Apply the previous Lemma for $S=A^{1/2}$ and $T=B^{1/2}$.
I've stolen the trick above from
a paper by Kato, where he shows a similar result for unbounded operators. I.e. when one defines $A\le B$ correctly (paying attention to domains), then $0<A\le B$ implies $0<B^{-1}\le A^{-1}$. This can be used to derive operator monotonicity of the maps $f(z)=z^\nu$ ($0\le \nu \le 1$), i.e. $0\le A \le B$ implies $f(A)\le f(B)$.
One more remark: It is not true in general that $\Vert a^{-1} \Vert = \Vert a \Vert^{-1}$. To see this, assume that $a_n$ is a sequence of invertible elements with limit $a_*\neq 0$. Then the sequence $\Vert a_n^{-1} \Vert = \Vert a_n \Vert^{-1}$ would be bounded by some $C<\infty$. Hence $\Vert a_n^{-1} - a_m^{-1}\Vert \le C^2 \Vert a_n - a_m \Vert \rightarrow 0$, which means that $a_n^{-1}$ must have a limit and it's easy to see that this must be an inverse of $a_*$. But this would imply that the set of invertible elements is open and closed in the set of nonzero elements and thus every nonzero $a$ was invertible, which is absurd unless the $C^*$-algebra is $\mathbb{C}$.
