Since the Kähler identities do not involve coordinates, it is natural to expect that there is a coordinate-free proof. However, it is completely unreasonable to expect that the coordinate-free proof is in any way more elegant.

Here is my rationale: if you read a proof using coordinates, and \it{this proof does not use any special property of the chosen coordinates}, then you can translate the argument into a coordinate-free one.

But the argument for the Kaehler identities is NOT: choose an arbitrary coordinate system and do a stupid calculation. If this were the case, Deligne, Sullivan, Griffiths and Morgan would surely have avoided coordinates. Instead, the argument is (in my opinion this argument is marvelous):

1. check the identities on $C^n$ with the standard metric;
2. at a given point $x \in M$, there is a coordinate system, mapping $x $ to $0$, such that the $1$-jet of the metric/complex structure is the ($1$-jet of the) standard metric/complex structure at $0$. Here Kählerness comes in, and if I remember correctly, exponential coordinates do the job.
3. Since the Kähler identities involve only the $1$-jet of the metric/complex structure and completely invariant things, the formulae are valid on a general Kähler manifold.

The standard proof of point 1 is a bit messy, and my suggestion is to make it slicker by using the symmetries of $C^n$. Observe that all expressions on $C^n$ are translation-invariant and invariant under $U(n)$. Moreover, the effect of scaling $C^n$ and conjugation is easy to figure out. These properties should be enough to force the Kähler identities.

In a completely coordinate-free proof, you do no longer have these special coordinates at hand, and thus I expect the proof, though possible, to be much more complicated.

Since the Kähler identities do not involve coordinates, it is natural to expect that there is a coordinate-free proof. However, it is completely unreasonable to expect that the coordinate-free proof is in any way more elegant.

Here is my rationale: if you read a proof using coordinates, and \it{this proof does not use any special property of the chosen coordinates}, then you can translate the argument into a coordinate-free one.

But the argument for the Kaehler identities is NOT: choose an arbitrary coordinate system and do a stupid calculation. If this were the case, Deligne, Sullivan, Griffiths and Morgan would surely have avoided coordinates. Instead, the argument is (in my opinion this argument is marvelous):

1. check the identities on $C^n$ with the standard metric;
2. at a given point $x \in M$, there is a coordinate system, mapping $x $ to $0$, such that the $1$-jet of the metric/complex structure is the ($1$-jet of the) standard metric/complex structure at $0$. Here Kählerness comes in, and if I remember correctly, exponential coordinates do the job.
3. Since the Kähler identities involve only the $1$-jet of the metric/complex structure and completely invariant things, the formulae are valid on a general Kähler manifold.

The standard proof of point 1 is a bit messy, and my suggestion is to make it slicker by using the symmetries of $C^n$. Observe that all expressions on $C^n$ are translation-invariant and invariant under $U(n)$. Moreover, the effect of scaling $C^n$ and conjugation is easy to figure out. These properties should be enough to force the Kähler identities.

In a completely coordinate-free proof, you do no longer have these special coordinates at hand, and thus I expect the proof, though possible, to be much more complicated.

EDIT: as RdN points out, Huybrechts gives a coordinate-free proof in his book. Huybrecht's proof is about as elegant and complicated as the standard proof.