Is always possible define a coordinate basis for a smooth manifold? The coordinate basis or holonomic basis for a differentiable manifold $\mathcal{M}$ is a set of basis vector fields $\{e_\mu\}$ definited in each point $P\in \mathcal{M}$ with the local condition
$$
[e_\mu,e_\nu]=\mathcal{L}_{e_\mu}e_\nu=0
$$
We also know that we can define a local non-coordinate basis as $\hat{e}_\mu=A^\nu_{\ \mu} e_\nu$, with $A\in GL(n,\mathbb{R})$ (in general).
But I was wondering if the viceversa is also true. For example, the Lie group $SO(3)$ (which is also a smooth manifold of course) has 3 generators $V_i$ and we know that structure constants are non-zero, so this isn't a coordinate basis, according to the previous definition. Is it possible to find a coordinate basis in this case? Does a manifold always have a coordinate basis?

I try to explain my doubt better, using the example of the Lie group $SO(3)$. Using the following non coordinate basis
$$
V_1=\pmatrix{0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0} \ \ \ \ \ V_2=\pmatrix{0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0} \ \ \ \ \ V_3=\pmatrix{0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0} \ \ \in \mathfrak{so}(3)
$$
the structure constants are
$$
[V_i,V_j]=\varepsilon_{ijk}V_k
$$
and we know that we can think at structure constants as something that completely define a Lie group. However, we are free to choose some other basis and we consequently obtain other structure constants, since they are nothing more that components of the vector $V=\varepsilon_{ijk}V_k$, so I expect that they appropriately change if I change the basis. But, if I can also use a coordinate base, than $V=0$ is the null vector, which is null in every basis, so in this case I can't go back to the old structure constants. I'm sorry if I said something wrong but I'm not well versed in this topic.
 A: Lukas has given a great answer, I just want to spell out the computation (about matrix Lie group).
Every matrix Lie group $G \subset GL(n, \mathbb R)$ is also a smooth manifold, thus there is actually two concepts of Lie bracket here. First you have
$$ \Gamma(G) \times \Gamma(G) \to \Gamma (G), \qquad (\tilde X, \tilde Y)\mapsto [\tilde X, \tilde Y]$$
here $\Gamma(G)$ denotes the smooth vector fields on $G$ and the bracket is the one defined by the smooth structure.
Second, there is another Lie bracket on the Lie algebra $\mathfrak g$:
$$\mathfrak g \times \mathfrak g\to \mathfrak g, \qquad (X, Y)\mapsto [X, Y]_0 = XY-YX,$$
here $\mathfrak g$ is a subspace of $M_n(\mathbb R)$ (spaces of $n\times n$ matrices) and $XY-YX$ is just the matrix multiplication.
Of course there is a way to link them together: there is a linear map
$$\mathfrak g \to \Gamma(G), \qquad X\mapsto \tilde X$$
so that
$$ \tag{1} [\tilde X, \tilde Y] = \widetilde{[X, Y]_0}, \qquad \forall X, Y \in \mathfrak g.$$
The map is defined as follows: $\mathfrak g$ is the tangent space of $G$ at the identity $e$. For each $A \in G$, the matrix multiplcation $X\mapsto AX$ defines isomorphism of tangent planes $T_eG \to T_A G$. Thus for each $X\in \mathfrak g$, the mapping
$$G\to TG, \qquad A \mapsto AX$$
is a smooth vector fields on $G$, which we define as $\tilde X$. With this definition we do have (1).
Going back to your confusion about using coordinates basis (and so can't recover the structural constants): you mixed those two concepts of Lie brackets. You can mix those two only when (1) is satisfied, or put it differently, only when the smooth coordinates vector fields happens to be $\tilde X$ for some $X\in \mathfrak g$. The fact that some structural constant are non-zero is telling you that there isn't a coordinates basis $\{V_1, \cdots, V_n\}$ so that $V_i = \tilde X_i$ for all $i$.
A: If you're given a differentiable manifold $\mathcal{M}$, like CyclotomicField said in his comment, by definition, around every point $p \in \mathcal{M}$ there is a chart around $p$, which basically means you have local coordinates $x_1,\dots,x_n$ around $p$. And then, with these local coordinates you can always define a corresponding coordinate basis $e_1,\dots,e_n$ which has this property $[e_\mu,e_\nu] = 0$ that you're looking for. So yes, around every point, you will find such a coordinate basis!
What you say about $SO(3)$ is then true: The Lie algebra elements $V_1,V_2,V_3 \in \mathfrak{g} = T_\text{id}G$ that you defined can be used to also define a basis of your Lie group $G$, but the Lie brackets between these vector fields do not vanish. And even if you change your basis in $\mathfrak{g}$ with some linear transformation $V_1 \mapsto V_1', V_2 \mapsto V_2', V_3 \mapsto V_3'$, no matter how you choose this transformation, the structure constants will not become zero, so no linear transformation will change this basis into a coordinate basis.
However, if you consider $V_1, V_2, V_3$ as a coordinate basis on $SO(3)$, you are not not simply seeing them as elements in $\mathfrak{g}$, but as vector fields on $SO(3)$. So you can do more than apply linear transformations with elements of $g_0 \in GL(3,\mathbb{R})$: you can perform local transformations with maps $g: SO(3) \to GL(3,\mathbb{R})$. This means that rather than applying a single rigid transformation on your vector field, you apply a different transformation of $GL(3,\mathbb{R})$ at every point of $SO(3)$. Performing such a transformation, you get that the transformed $V_i'$ has the shape
$$V_i(x) = \sum_{i=1}^3 g_{ij}(x) V_i(x).$$
Doing this, you can actually manage to "bend" your basis $V_1, V_2, V_3$ to get a commuting coordinate basis $V_1',V_2',V_3'$ if you choose your transformation function $g$ appropriately. The Lie brackets between the $V_i'$ then also depend on the derivatives of $g$, and that is the necessary degree of freedom that you need to make things commute.
