Is it possible to have renormalisable C-, P-, or T-violating terms in QED?

Although I'm not sure about a general proof of the $C$, $P$, and $T$ invariance of QED, I can explain why OP's examples don't invalidate Witten's argument.

(1) $m\bar{\psi} \psi + m_5 \bar{\psi} \gamma_5\psi$ is unitarily equivalent to $\sqrt{m^2+m_5^2}\,\bar{\psi} \psi$, where the relevant unitary operation is of the form $e^{\theta\gamma_5}$. Here, $\theta$ is a real parameter, and notice that this operation leaves the kinetic term invariant. The parity and charge conjugation symmetries hold in modified forms, that is, the corresponding symmetry operations are $P_{\mathrm{new}} := e^{\theta\gamma_5} P e^{-\theta\gamma_5}$ and $C_{\mathrm{new}} := e^{\theta\gamma_5} C e^{-\theta\gamma_5}$.

(Note that OP's definition of $\gamma_5$ is missing a factor of $i$ from the most widely used one, i.e., $i\gamma_0\gamma_1\gamma_2\gamma_3$. As a consequence, $\gamma_5$ is an anti-Hermitian matrix here.)

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Below are no longer valid. OP meant $m_c\psi^T C\gamma^0 \psi$ as a possible mass term, but in the first version of the post, the term was written as $m_c \bar{\psi}C\psi$. Discussions below are referring to the first version of the post.

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(2) The term $m_c \bar{\psi}C\psi$ is not a mass term. To see this, consider the following Lagrangian: \begin{equation} \mathcal{L} = i\bar{\psi}\gamma^\mu\partial_\mu \psi + m\bar{\psi}\psi + im_c \bar{\psi}C\psi, \end{equation} where $m$ and $m_c$ are real parameters. The corresponding Hamiltonian is \begin{equation} \mathcal{H} = -i\psi^\dagger\gamma^0\gamma^i \partial_i\psi - m\psi^\dagger\gamma^0\psi - im_c \psi^\dagger\gamma^2\psi. \end{equation} (Be reminded that $\bar{\psi} = \psi^\dagger\gamma^0$.) Here, the usual mass term $m\psi^\dagger\gamma^0\psi$ anticommutes with the kinetic term $-i\psi^\dagger\gamma^0\gamma^i \partial_i\psi$, which is the origin of the Dirac spectrum $\pm\sqrt{p^2 + m^2}$.

However, $im_c \psi^\dagger\gamma^2\psi$ anticommutes with the conventional mass term and the $i=2$ part of the kinetic term, while commuting with the rest of the kinetic term. As a result, the spectrum of the above Hamiltonian is given by \begin{equation} E(\textbf{p}) = \pm\sqrt{\Big(\sqrt{p_1^2 + p_3^2}\pm m_c\Big)^2 + p_2^2 + m^2}. \end{equation} Here, $m_c$ obviously doesn't play the role of a mass gap, and even more, it makes the $y$ direction inequivalent from the $x$ and $z$ directions, breaking Lorentz invariance.

Update: In part (2), writing down the explicit spectrum certainly wasn't the best way to demonstrate why $im_c \psi^\dagger\gamma^2\psi$ cannot be a valid term in a QED Lagrangian. A better way would be to note that this term, unlike other terms in the Hamiltonian, does not commute with all three rotation generators \begin{equation} J^{ij} = -i(x^i\partial_j - x^j\partial_i) + \frac{i}{4}[\gamma^i , \gamma^j]. \end{equation} To be specific, $im_c \psi^\dagger\gamma^2\psi$ commutes with $J^{31}$ but does not with $J^{12}$ or $J^{23}$. Therefore, the Hamiltonian is invariant under a rotation in the 3-1 plane but not in the 1-2 or 2-3 plane. This fact, indeed, is reflected in the above spectrum.