Why is left multiplication on a group bijective?

The map $\ell_g$ defined by $\ell_g(h)=gh$ is a bijection $G\to G$. Indeed, note that the map $\ell_{g^{-1}}=\ell_g^{-1}$ because $$ (\ell_{g^{-1}}\circ\ell_g)(h)=g^{-1}gh=h$$ and similarly $$ (\ell_g\circ \ell_{g^{-1}})(h)=gg^{-1}h=h.$$ The analogous argument shows that the right multiplication operator is a bijection. Note that none of these operators is a group homomorphism except for example $\ell_e=\Bbb{1}_G.$

Suppose $gx=gy$. Then $g^{-1} gx=g^{-1} gy \implies x=y$. That proves injectivity.

Suppose $g'\in G$. Then $g (g^{-1}g')=g'$ implies surjectivity.

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In addition, in terms of actions, left multiplication by an element of a group is transitive and faithful. These are somewhat similar notions.