As your question already hints to, the "extra object" preserved is the Lie bracket. I.e., for any connected semisimple Lie group $G$, the image of $\mathrm{Ad}: G\to\mathrm{GL}(\mathfrak g)$ is precisely the identity component $\mathrm{Aut}(\mathfrak g)^0$ of $$ \mathrm{Aut}(\mathfrak g) = \{a\in\mathrm{GL}(\mathfrak g):a([X,Y])=[a(X),a(Y)] \text{ for all } X,Y\in\mathfrak g\}. $$

Proof:

Clearly $\mathrm{Ad}$ maps $G$ into that identity component. To see that it is onto it, it is enough to see that its derivative $\mathrm{ad}:\mathfrak{g}\to\mathrm{gl}(\mathfrak g)$ is onto the Lie algebra $$ \mathrm{aut}(\mathfrak g) = \{D\in\mathrm{gl}(\mathfrak g):D([X,Y])=[D(X),Y])+[X,D(Y)] \text{ for all } X,Y\in\mathfrak g\}, $$ or in other words, that every derivation $D$ of $\mathfrak g$ is inner: $D=\mathrm{ad}(Z)$ for some $Z\in\mathfrak g$. But that is precisely the content of Whitehead's Lemma (cf. any book on semisimple Lie algebras).

As your question already suggests, the "extra object preserved" is the Lie bracket. I.e., for any connected semisimple Lie group $G$, the image of $\mathrm{Ad}: G\to\mathrm{GL}(\mathfrak g)$ is precisely the identity component $\mathrm{Aut}(\mathfrak g)^0$ of $$ \mathrm{Aut}(\mathfrak g) = \{a\in\mathrm{GL}(\mathfrak g):a([X,Y])=[a(X),a(Y)] \text{ for all } X,Y\in\mathfrak g\}. $$

Proof. Clearly $\mathrm{Ad}$ maps $G$ into that identity component. To verify that it is onto it, it is enough to see that its derivative $\mathrm{ad}:\mathfrak{g}\to\mathrm{gl}(\mathfrak g)$ is onto the Lie algebra $$ \mathrm{aut}(\mathfrak g) = \{D\in\mathrm{gl}(\mathfrak g):D([X,Y])=[D(X),Y]+[X,D(Y)] \text{ for all } X,Y\in\mathfrak g\} $$ of $\mathrm{Aut}(\mathfrak g)$, or in other words, that every derivation $D$ of $\mathfrak g$ is inner: $D=\mathrm{ad}(Z)$ for some $Z\in\mathfrak g$. But that is precisely the content of Whitehead's Lemma (see any book on semisimple Lie algebras).

For any semisimple Lie group $G$, the image of $\mathrm{Ad}: G\to\mathrm{GL}(\mathfrak g)$ is the identity component $\mathrm{Aut}(\mathfrak g)^0$ of $$ \mathrm{Aut}(\mathfrak g) = \{a\in\mathrm{GL}(\mathfrak g):a([X,Y])=[a(X),a(Y)] \text{ for all } X,Y\in\mathfrak g\}. $$

As your question already hints to, the "extra object" preserved is the Lie bracket. I.e., for any connected semisimple Lie group $G$, the image of $\mathrm{Ad}: G\to\mathrm{GL}(\mathfrak g)$ is precisely the identity component $\mathrm{Aut}(\mathfrak g)^0$ of $$ \mathrm{Aut}(\mathfrak g) = \{a\in\mathrm{GL}(\mathfrak g):a([X,Y])=[a(X),a(Y)] \text{ for all } X,Y\in\mathfrak g\}. $$

Proof:

Clearly $\mathrm{Ad}$ maps $G$ into that identity component. To see that it is onto it, it is enough to see that its derivative $\mathrm{ad}:\mathfrak{g}\to\mathrm{gl}(\mathfrak g)$ is onto the Lie algebra $$ \mathrm{aut}(\mathfrak g) = \{D\in\mathrm{gl}(\mathfrak g):D([X,Y])=[D(X),Y])+[X,D(Y)] \text{ for all } X,Y\in\mathfrak g\}, $$ or in other words, that every derivation $D$ of $\mathfrak g$ is inner: $D=\mathrm{ad}(Z)$ for some $Z\in\mathfrak g$. But that is precisely the content of Whitehead's Lemma (cf. any book on semisimple Lie algebras).