Evolution of density operators under decoherence
In terms of density operators, the qubit alone evolves from the pure state |\psi\rangle\langle\psi| to a mixed state, which can be obtained by tracing over the environment.
We start with the evolution of the state vector |\psi\rangle=\alpha|0\rangle+\beta|1\rangle, which is given by
\left( \alpha|0\rangle +\beta |1\rangle\right)|e\rangle \longmapsto
\alpha |0\rangle|e_{00}\rangle +\beta |1\rangle |e_{11}\rangle,
Then we write it as the evolution of the projector |\psi\rangle\langle\psi|, and trace over the environment to obtain
\begin{aligned}
|\psi\rangle\langle\psi| \longmapsto & |\alpha|^2|0\rangle\langle 0| \langle e_{00}|e_{00}\rangle+ \alpha\beta^\star |0\rangle\langle 1|\langle e_{11}|e_{00}\rangle
\\+ &\alpha^\star\beta |1\rangle\langle 0|\langle e_{00}|e_{11}\rangle + |\beta|^2|1\rangle\langle 1|\langle e_{11}|e_{11}\rangle.
\end{aligned}
Written in the matrix form, this is
\begin{bmatrix}
|\alpha|^2 & \alpha\beta^\ast
\\\alpha^\ast\beta & |\beta|^2
\end{bmatrix}
\longmapsto
\begin{bmatrix}
|\alpha|^2 & \alpha\beta^\ast \langle e_{11}|e_{00}\rangle
\\\alpha^\ast\beta \langle e_{00}|e_{11}\rangle & |\beta|^2
\end{bmatrix}.
The off-diagonal elements, originally called coherences, vanish as \langle e_{00}|e_{11}\rangle approaches zero.
This is why this particular interaction is called decoherence.
Notice that
|\psi\rangle|e\rangle \longmapsto \mathbf{1}|\psi\rangle|e_{\mathbf{1}}\rangle+Z|\psi\rangle|e_Z\rangle,
implies
|\psi\rangle\langle\psi|\longmapsto \mathbf{1}|\psi\rangle\langle\psi| \mathbf{1}\langle e_{\mathbf{1}}|e_{\mathbf{1}}\rangle +Z|\psi\rangle\langle\psi| Z\langle e_Z|e_Z\rangle,
only when \langle e_{\mathbf{1}}|e_Z\rangle=0 (otherwise you would have additional cross terms \mathbf{1}|\psi\rangle\langle\psi| Z and Z|\psi\rangle\langle\psi| \mathbf{1}).
In this case we can indeed say that, with probability \langle e_{\mathbf{1}}|e_{\mathbf{1}}\rangle, nothing happens, and, with probability \langle e_Z|e_Z\rangle, the qubit undergoes the phase-flip Z.