The map $G \times G \rightarrow G$ defined as $(x,y) \mapsto xy$ is differentiable. The left translation is $L_x(y) = xy$. To show that it is a diffeomorphism, we need it to be a bijection, that is both differentiable and has a differentiable inverse.
It is injective because if $xy_1 = xy_2$ then $y_1=y_2$.
It is surjective because for some arbitrary element $z \in G$ we can multiply on the left by $x^{-1}$ to get $x^{-1}z=y$. Then $L_x(y)=z$, meaning that $z\in \operatorname{im}(L_x)$.
But why is $L_x$ differentiable, and why is its inverse?