This is a proof using just the definition of a Lie group. Let $m: G\times G \to G$ be the multipication, then $$ dm_e : T_{(e,e)} (G\times G) \cong T_eG \oplus T_eG \to T_e G$$ is given by $(X, Y) \mapsto X+Y$. This is true as clearly $dm_e (X, 0) =X$ and $dm_e (0, Y) = Y$ and that $dm_e$ is linear.

Now the composition

$$ G \overset{(\operatorname{id}, i)}{\to} G\times G \overset{m}{\to}G$$ is the constant map to $e$, using the Chain rule we have (for all $X\in T_eG$)

$$ 0= dm_e ( \operatorname{id}_* X, i_* X) = X + i_*X,$$

thus $i_*X = -X$.

This is a proof using just the definition of a Lie group. Let $m: G\times G \to G$ be the multiplication, then $$ dm_e : T_{(e,e)} (G\times G) \cong T_eG \oplus T_eG \to T_e G$$ is given by $(X, Y) \mapsto X+Y$. This is true as clearly $dm_e (X, 0) =X$, $dm_e (0, Y) = Y$ and that $dm_e$ is linear.

The composition

$$ G \overset{(\operatorname{id}, i)}{\to} G\times G \overset{m}{\to}G$$ is constant (everything maps to $e$). Using the chain rule we have (for all $X\in T_eG$)

$$ 0= dm_e ( \operatorname{id}_* X, i_* X) = X + i_*X,$$

thus $i_*X = -X$.