A few observations:
Saying that an "entanglement breaking (EB) channel describes the measurement process" is just not a very meaningful statement. First of all, it being a single channel as I defined it, it can't be describing post-measurement states.
It could possibly describe the post-measurement state when we don't know the outcome. However, that is not the case in general, as explained by the other answer and as I will explicitly show in my examples below.
The most general way to describe measurements with their post-measurement states is via quantum instruments: channels of the form $\Phi(\rho)=\sum_i \Phi_i(\rho)\otimes \mathbb{P}_i$ with $\Phi_i$ completely positive maps such that $\sum_i \Phi_i$ is CPTP.
These can give rise to entanglement breaking channel in the sense that $\sum_i \Phi_i$ might be EB. This is typically the case when the measurement is "maximally destructive".
These channels are also connected to the POVM $\mu$ describing the same measurement via $\mu_i= \Phi_i^\dagger(I)$.
This is also discussed in this other answer.
How destructive a measurement is determins how reversible or close to reversible the channels describing post-measurement states can be.
E.g. in the extreme case of a trivial $m$-outcome POVM, the quantum instrument needs to have $\Phi_i^\dagger(I)=I/m$, which can be satisfied by $\Phi_i(\rho)=\frac1m U_i\rho U_i^\dagger$ for any set of (not necessarily distinct) unitaries $U_i$. In such situation the act of measurement is perfectly reversible (and in the limit case of $U_i=U_j$ there's no measurement feedback at all).
It's still possible to implement the measurement in other ways that are not reversible, but that's unavoidable, see again this answer for more details on this point.
We can also characterise the other end of the spectrum: consider a rank-1 POVM, with elements $\mu_a = w_a \mathbb{P}_{\psi_a}$ with $w_a>0$ such that $\sum_a w_a= d$, with $d$ dimension of the space. That means $\Phi_a^\dagger(I)=w_a \mathbb{P}_{\psi_a}$, which in turn implies that $\Phi_a^\dagger$ must be a replacement map: $\Phi_a^\dagger(\rho) = \operatorname{tr}(\rho)w_a\mathbb{P}_{\psi_a}$, and thus also $\Phi_a(\rho) = w_a I \langle\psi_a|\rho|\psi_a\rangle$.
In other words, all post-measurement states must be the equal to each other, and to the maximally mixed state.
Here's a few explicit examples that I think clarify the issue:
Example with projective POVM
Consider a two-qubit system, where we want to measure a single-qubit input in the first register. To perform this measurement, we perform a unitary evolution through a Hadamard gate on the first qubit followed by a CNOT with the second qubit as target, and then measure the second register on the computational basis.
If the second qubit starts in $|0\rangle$, and the input qubit in the first register is $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, then the output is (neglecting normalisation):
$$|\psi\rangle|0\rangle\to (\alpha+\beta)|00\rangle+(\alpha-\beta)|11\rangle.$$
It follows that measuring the second qubit in the computational basis completely collapses the first qubit.
The POVM $\mu\equiv \{\mu_0,\mu_1\}$ describing the situation is projective with elements $\mu_0=\mathbb{P}_+$ and $\mu_1=\mathbb{P}_-$.
The state of the first register ignoring the measurement outcome can be described by an entanglement breaking channel:
$$\Phi(\rho) = \langle \mathbb{P}_+,\rho\rangle \mathbb{P}_0 + \langle\mathbb{P}_-,\rho\rangle \mathbb{P}_1,
\qquad \mathbb{P}_\psi\equiv |\psi\rangle\!\langle\psi|.$$
We can characterise both post-measurement states and outcome probabilities with a quantum instrument as:
$$\Phi_{\rm QI}(\rho) = \Phi_0(\rho)\otimes \mathbb{P}_0+\Phi_1(\rho)\otimes \mathbb{P}_1,
\quad \Phi_0(\rho)\equiv \langle\mathbb{P}_{+},\rho\rangle \mathbb{P}_{0},
\quad
\Phi_1(\rho)\equiv \langle\mathbb{P}_{-},\rho\rangle \mathbb{P}_{1}.$$
Note that starting from the POVM $\mu$, we could have obtained the Kraus operators $A_0=|0\rangle\!\langle +|$ and $A_1=|1\rangle\!\langle -|$, and written the post-measurement states as $\Phi_i(\rho)= A_i \rho A_i^\dagger$.
The POVM is then successfully recovered from these maps as $\Phi_i^\dagger(I)=\mu_i$.
Example with trivial two-outcome POVM
On the other hand, suppose after the same evolution the second register is measured in the eigenbasis of $\sigma_X$ instead. Or equivalently, say we're applying a Hadamard gate to the second register before measuring in the computational basis. The evolution then reads:
$$|\psi\rangle|0\rangle \to [(\alpha+\beta)|0\rangle+(\alpha-\beta)|1\rangle]\otimes |0\rangle
+
[(\alpha+\beta)|0\rangle-(\alpha-\beta)|1\rangle]\otimes |1\rangle.$$
In this case the outcome probabilities are $\|(\alpha+\beta)|0\rangle\pm(\alpha-\beta)|1\rangle\|^2=1/2$, and thus the corresponding POVM is the trivial one: $\mu_0=\mu_1=I/2$.
In other words, this measurement doesn't "steal" any amount of information from the input state.
As a quantum instrument this measurement is described by
$$\Phi_{\rm QI}(\rho) = \sum_{i=0,1}\Phi_i(\rho)\otimes\mathbb{P}_i, \\
\Phi_i(\rho) \equiv \frac14(I \langle I,\rho\rangle + Z \langle X,\rho\rangle \pm X \langle Z,\rho\rangle \mp Y\langle Y,\rho\rangle).$$
In fact, one can also just observe directly from the output state that $\Phi_i$ are unitary channels up to weight factors: $\Phi_i(\rho)=\frac12 U_i \rho U_i^\dagger$ with
$$U_0 = H\equiv \frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix},
\quad
U_1 = ZH= \frac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}.$$
This is not accidental: if the measurement is trivial, no information is stolen from the input state, which means the effect of the measurement is reversible, i.e. unitary.
Finally, ignoring the outcome the post-measurement state is
$$\Phi(\rho) = \frac12(U_0\rho U_0^\dagger + U_1 \rho U_1^\dagger),$$
which is a mixed-unitary channel. This channel is again entanglement-breaking, as its Choi is
$J(\Phi) = \mathbb{P}_0\otimes\mathbb{P}_++\mathbb{P}_1\otimes\mathbb{P}_-$ (see also here about EB channels).
However, the same identical POVM, $\mu_i=I/2$, can correspond to a quantum instrument which gives a non-EB post-measurement outcome-agnostic state.
This is borderline trivial: the trivial measurement can be implemented by having the second register not interact with the input state at all. In which case the instrument would just be $\Phi_{\rm QI}(\rho) = \frac12\rho+\frac12\rho=\rho$, which is also the same as the state you get ignoring the measurement outcome.
And of course, the identity channel is not entanglement breaking.
Example with a partially depolarising channel
As shown here, the depolarising channel for a single qubit, $\Phi_p(\rho)=p\rho+(1-p)\operatorname{tr}(\rho) I/2$, can be implemented via the Stinespring isometry:
$$V = \begin{pmatrix}
\sqrt{\frac{1+3p}{4}} \,I \\
\sqrt{\frac{1-p}{4}} \,Z \\
\sqrt{\frac{1-p}{2}} E_{01} \\
\sqrt{\frac{1-p}{2}} E_{10}
\end{pmatrix}
= \sum_{i=1}^4 |i\rangle\otimes A_i.$$
This isometry acts on a three-qubit space, with the first qubit being our input, and two ancillary qubits needed to obtain the channel.
Let's consider a situation where we first evolve the state through this isometry $V$, and then measure in the computational basis the last two qubits.
We thus have four possible outcomes, and conditionally to observign the $i$-th outcome the first qubit becomes, up to renormalisation, $\Phi_i(\rho)\equiv A_i \rho A_i^\dagger$. More explicitly, we have
$$\Phi_1(\rho) = \frac{1+3p}{4} \rho,
\qquad
\Phi_2(\rho) = \frac{1-p}{4} Z\rho Z,
\\
\Phi_3(\rho) = \frac{1-p}{2} \mathbb{P}_0 \langle\mathbb{P}_1,\rho\rangle,
\qquad
\Phi_4(\rho) = \frac{1-p}{2} \mathbb{P}_1 \langle\mathbb{P}_0,\rho\rangle.$$
The corresponding POVM is then $\mu_i=\Phi_i^\dagger(I)$, that is,
$$
\mu_1 = \frac{1+3p}{4} I,
\quad
\mu_2 = \frac{1-p}{4} I,
\quad
\mu_3 = \frac{1-p}{2} \mathbb{P}_1,
\quad
\mu_4 = \frac{1-p}{2} \mathbb{P}_0.
$$
The channel describing the state when we do not know the measurement outcome is $\Phi=\sum_i \Phi_i$, which here is by construction nothing but the original depolarising channel $\Phi_p$.
It therefore follows that in this case the post-measurement state is not given by an entanglement breaking channel, except for the trivial case $p=0$.
