For any element $g\in G$ we have a left multiplication map $L_g\colon G\to G$. Given $x\in T_1G$, we can define $X\in \Gamma(TG)$ by \begin{equation} X(g) = dL_g(x). \end{equation} Notice that $X(1)=dL_1(x)=x$, since $L_1$ is the identity map on $G$. Also, $X$ is left-invariant by construction. Now suppose $Y\in\Gamma(TG)$ is left invariant and $Y(1)=x$. Then for any $g\in G$, \begin{equation} Y(g)=dL_g(Y(1))=dL_g(x)=X(g), \end{equation} so $Y=X$, meaning that $X$ is the unique left-invariant vector field with $X(1)=x$.