Arbitrary error correctable iff Pauli error are: misunderstanding from Preskill notes

I am following Preskill notes.

What I want to understand is why it is in general enough to be able to correct n-qubit Pauli errors to say that an arbitrary error can be corrected.

I call: $mathcal{M}(rho)=sum_a M_a rho M_a^{dagger}$ an error map, $C$ the code space.

A CPTP recovery operation exists if and only if, the Kraus operator of the error map verify the Knill-Laflamme condition:

$$forall (i,j) in C : langle overline{i} | M_{delta}^{dagger} M_{mu} | overline{j} rangle = C_{delta mu} delta_{ij}$$

With $C_{delta mu}$ an Hermitian matrix and $langle overline{i} | overline{j} rangle=delta_{ij}$ (the family $| overline{i} rangle$ forms an orthonormal basis of $C$).

My question

From this condition, and the fact any Kraus operator can be decomposed as a sum of n-qubit Pauli matrices, Preskill seem to say that it shows that if one is able to correct against Pauli error it is able to correct against an arbitrary error (that verifies Knill Laflamme condition of course).

I call $mathcal{E}$ the set of n-qubit Pauli operators on which we can decompose any of the $M_a$‘s.

He says on page 10, just below (7.26)

"since each $E_a$ in $mathcal{E}$ is a linear combination of $M_{mu}$‘s, then"

$$forall (i,j) in C: langle overline{i} | M_{delta}^{dagger} M_{mu} | overline{j} rangle = C_{delta mu} delta_{ij} Rightarrow forall (i,j) in C: langle overline{i} | E_{a}^{dagger} E_{b} | overline{j} rangle = C_{ba} delta_{ij}$$

And I don’t understand this. I could imagine $mathcal{M}(rho)=U rho U^{dagger}, U notin mathcal{E}$ (a unitary error that is not strictly a Pauli one). Wouldn’t this be a counter example in which the Pauli matrix on which the error can be decomposed cannot in turn be expressed in function of Kraus operator.

To say things more shortly: is there an easy way to show that if one is able to correct Pauli error we can show it is equivalent to Knill Laflamme condition?