Show that $\langle S^{-1}\rangle=\langle S\rangle$. In particular, $\langle a^{-1}\rangle=\langle a\rangle$, so also $o(a) = o(a^{-1})$

Your attempt is fine. It seems like you understand the question.

Since, for $S=\{ s_i\mid i\in I\}$ for some index set $I$,

$$\langle S\rangle =\left\{ s_{i_1}^{\varepsilon_{i_1}}\dots s_{i_k}^{\varepsilon_{i_k}}\,\middle|\, i_j\in I, k\in\Bbb N, \varepsilon_{i_j}\in\{1,-1\}\right\}\cup\{ e\},$$

we have

$$\langle S^{-1}\rangle =\left\{ s_{i_1}^{\delta_{i_1}}\dots s_{i_k}^{\delta_{i_k}}\,\middle|\, i_j\in I, k\in\Bbb N, \delta_{i_j}\in\{1,-1\}\right\}\cup\{ e\},$$

where $\delta_{i_j}=-\varepsilon_{i_j}$ implies $\langle S\rangle=\langle S^{-1}\rangle$; so if we let $S=\{a\}$, then by definition we get

$$\begin{align} \langle a\rangle &=\langle \{ a\}\rangle \\ &=\langle \{ a\}^{-1}\rangle\\ &=\langle \{ a^{-1}\}\rangle \\ &=\langle a^{-1}\rangle. \end{align}$$

Moreover, it is fairly simple (if not tautological by definition) that $o(x)=\lvert\langle x\rangle\rvert$ for all $x\in G$. Thus

$$\begin{align} o(a)&=\lvert \langle a\rangle\rvert\\ &=\lvert \langle a^{-1}\rangle\rvert\\ &=o(a^{-1}). \end{align}$$