However, the following is confusing:
In a reference frame that sees the protons moving, the same forces
result from the magnetic force created by the protons' movement.
I agree that it is confusing. In the instant that the two protons begin to move away from each other they are momentarily mutually at rest. At this instant, in a frame of reference in which the proton's are (momentarily) at rest, the force on each proton due to the other is purely electric in nature.
At this instant, in this frame, the coordinate acceleration of either proton is equal to its proper acceleration (the acceleration as measured by an 'accelerometer' on the proton)
However, in the lab frame, special relativity dictates that the coordinate acceleration of the protons away from each other $(\mathbf{a}'_\perp)$ is actually less than the proper acceleration away from each other $(\boldsymbol{\alpha}_\perp)$:
$$\mathbf{a}'_\perp = \frac{1}{\gamma^2}\boldsymbol{\alpha}_\perp$$
But we also have that
$$\mathbf{F}'_\perp = \gamma m_p \mathbf{a}'_\perp$$
$$\mathbf{F}_\perp = m_p \boldsymbol{\alpha}_\perp$$
Thus, it must be that
$$\mathbf{F}'_\perp = \frac{1}{\gamma}\mathbf{F}_\perp$$
That is, in the lab frame, there is less of a repulsive force on each proton.
Since charge is Lorentz invariant, the electric force of repulsion between the protons is unchanged from the (momentary) rest frame and so, if there were only an electric force, there would be an inconsistency.
However, there is also an attractive magnetic force component in the lab frame and so, the electromagnetic force of repulsion between the protons is less in the lab frame and is thus consistent with SR.
To be sure, let's check the calculation. The fields at the location of the 'upper' proton due to the 'lower' proton in the momentary rest frame transform to the lab frame as
$$\begin{align}
& \mathbf {{E}_{\parallel}}' = \mathbf {{E}_{\parallel}}\\
& \mathbf {{B}_{\parallel}}' = \mathbf {{B}_{\parallel}} = 0\\
& \mathbf {{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right) = \gamma \mathbf {E}_{\bot} = \gamma \frac{e}{4\pi\epsilon_0d^2}\hat{\mathbf{z}}\\
& \mathbf {{B}_{\bot}}'= \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right) = -\gamma \frac{1}{c^2} \mathbf{ v} \times \mathbf {E} = +\gamma \frac{v}{c^2}\frac{e}{4\pi\epsilon_0d^2}\hat{\mathbf{x}}
\end{align}$$
where the lab frame has velocity $\mathbf{v} = -v\hat{\mathbf{y}}$ in the momentary rest frame.
Since the protons have zero velocity in the momentary rest frame, the Lorentz force on the 'upper' proton is
$$\mathbf{F} = e\left(\mathbf{E} + \mathbf{0} \times \mathbf{B} \right) = \frac{e^2}{4\pi\epsilon_0d^2}\hat{\mathbf{z}}$$
In the lab frame, the velocity of the protons is $\mathbf{u} = v\hat{\mathbf{y}}$ and the Lorentz force is
$$\mathbf{F}' = e\left(\mathbf{E}' + \mathbf{u} \times \mathbf{B}' \right)\frac{e^2}{4\pi\epsilon_0z^2} \gamma\left(1 - \frac{v^2}{c^2} \right)\hat{\mathbf{z}} = \frac{1}{\gamma}\mathbf{F}$$
as desired.
