Yes this is indeed exactly what is meant!
Perhaps unnecessary, but let me say what happens here in a more word-y, less notation-y way. The 'left multiplication'-action of $G$ on itself maps points of $G$ to points of $G$. 'Hence' it also maps tangent vectors to tangent vectors. As my physicist friends like to say: 'just like how a potato field is at every point a potato a vector field is at every point [of $G$] a vector'. So from the above it is obvious that the left action also maps vector fields to vector fields: viewing $X$ as a collection of vectors, the image of $X$ under a particular left-multiplication map $L_x$ consists just of all the images of all these vectors.
Being left-invariant means that as a collection of vectors, $X$ is mapped to itself under each $L_x$. Sure, individual vectors are mapped all over the place, but the total collection stays the same. In particular, if some $L_x$ maps $m$ to $g$ then the value of $X$ at $g$ is determined by the value of $X$ at $m$ since the vector in the tangent space at $g$ that $X(m)$ is mapped to, must belong to $X$ and $X$ only has one vector in that space. Very nice. Turning our heads 180 degrees we conclude that the last sentence means that $X(m)$ determines the value $X(g)$ for all $g$ for which there is some $x$ such that $L_xm = g$.
What then remains to be shown is that for each $g$ there exists an $x$ such that $L_xm = g$ and you achieved that marvelously by setting $x = gm^{-1}$.
EDIT: Now we also see what the more general case looks like. Instead of the smooth manifold being the group itself we might take it to be any manifold $M$ on which $G$ acts smoothly by a left-action $L$. In this case one would speak of $L$-invariant rather than left-invariant (except in sentences where left is used as a verb, like '$X$ is left invariant by $L$').
Anyway, the point is that if and only if the action of $G$ on $M$ is transitive, the same property holds, with the same proof. When the action is not necessarily transitive then the value of $X$ on each $G$-orbit is determined by the value at a single point (in that orbit).
Let $G$ be a Lie group and ${\cal G}$ its Lie algebra. Suppose that $e_1,...,e_n$ is a basis of ${\cal G}$, you can defined the left-invariant vector field $X_i(g)=(dL_g)_I(e_i)$, $(X_1(g),...,X_n(g))$ is a basis of the tangent space $T_gG$. Let $(a_1,...,a_n)$ be the dual basis of $(e_1,...,e_n)$. Let $\alpha^i$ be the $1$-form defined by $\alpha^i_g(u)=a_i(dL_g^{-1}(u))$. For every vector field $X$, $X(g)=\alpha^1_g(X)X_1+...+\alpha^n_g(X)X_n$.
Best Answer
Let $G$ be a Lie group and $X$ a vector field. For $g\in G$, let $L_g:G\to G$ denote left-multiplication, i.e., $L_g(p)=gp$. Then we say that $X$ is left-invariant if $$(L_g)_*X=X.$$ Note that the pushforward makes sense on vector fields since $L_g$ is a diffeomorphism. Making this explicit at a point, say at $p\in G$ yields $$X_{L_g(p)}=((L_g)_*X)_p$$ (this is saying $X$ is $L_g$-related with itself). Or rewritten, we have that $$X_{gp}=d(L_g)_p(X_p).$$