Derive $\frac{\mathrm{d}}{\mathrm{d}t}(\gamma m\mathbf{v}) = e\mathbf{E}$ from elementary principles? It is experimentally known that the equation of motion for a charge $e$ moving in a static electric field $\mathbf{E}$ is given by:
$$\frac{\mathrm{d}}{\mathrm{d}t} (\gamma m\mathbf{v}) = e\mathbf{E}$$
Is it possible to show this using just Newton's laws of motion for the proper frame of $e$, symmetry arguments, the Lorentz transformations and other additional principles?

If I understand your question correctly, you can show it by the next way.
First of all, use some expressions from Special relativity, which are Lorentz transformations for force $\mathbf F $, radius-vector $\mathbf r$ and speed $\mathbf v$ ($\mathbf u$ is the speed of inertial system):
$$
\mathbf r' = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} - \gamma \mathbf u t = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} \quad(t = 0) \quad \Rightarrow r'^{2} = r^{2} + \frac{(\mathbf u \cdot \mathbf r)^{2}\gamma^{2}}{c^{2}},
$$
$$
(\mathbf u \cdot \mathbf r') = \gamma (\mathbf u \cdot \mathbf r), \quad (\mathbf v' \cdot \mathbf r') = \frac{(\mathbf r \cdot \mathbf v)}{\gamma (1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} - \gamma (\mathbf r \cdot \mathbf u),
$$
$$
\frac{\mathbf F}{\gamma(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} = \mathbf F' + \gamma \frac{\mathbf u (\mathbf F' \cdot \mathbf v')}{c^{2}} + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf F')}{c^{2}} \qquad (.1).
$$
Secondly, use (.1) for Coulomb's law. You can make it, because it doesn't have an information about speed of interaction, which can be proved by thought experiment with two charges, binded by the stiff spring at the rest state.
So,
$$
\frac{\mathbf F}{\gamma(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} = \frac{Qq}{r'^{3}}\!{\left[\mathbf r' + \gamma \mathbf u \frac{(\mathbf r' \cdot \mathbf v')}{c^{2}} + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r')}{c^{2}}\right]} = $$
$$ = \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} + \gamma \frac{\mathbf u}{c^{2}} \left(\frac{(\mathbf u \cdot \mathbf r)}{\gamma(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})} - \gamma (\mathbf u \cdot \mathbf r)\right) + \Gamma \gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}}\right]} = 
$$
$$
= \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}}(1 + \gamma) - \frac{\mathbf u}{c^{2}}\gamma^{2}(\mathbf u \cdot \mathbf r) + \gamma \frac{\mathbf u}{c^{2}}\frac{(\mathbf v \cdot \mathbf r)}{\gamma(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})}\right]} = | \Gamma (1 + \gamma) = \gamma^{2} | =
$$
$$
= \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \gamma^{2} \frac{\mathbf u}{c^{2}}(\mathbf u \cdot \mathbf r) - \frac{\mathbf u}{c^{2}}\gamma^{2}(\mathbf u \cdot \mathbf r)) + \gamma \frac{\mathbf u}{c^{2}}\frac{(\mathbf v \cdot \mathbf r)}{\gamma(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})}\right]} =
$$
$$
= \frac{Qq}{r'^{3}}\!{\left[\mathbf r + \frac{\mathbf u (\mathbf v \cdot \mathbf r)}{c^{2}(1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}})}\right]} = |\mathbf u (\mathbf v \cdot \mathbf r) = \left[ \mathbf v [\mathbf u \times \mathbf r ] \right] + \mathbf r (\mathbf u \cdot \mathbf v)| = 
$$
$$
= \frac{Qq}{r'^{3}}\left[ \mathbf r + \frac{[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}]}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}} + \frac{\mathbf r \frac{(\mathbf u \cdot \mathbf v)}{c^{2}}}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right] = 
$$
$$
= \frac{Qq}{r'^{3}}\left[ \frac{[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}]}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}} + \frac{\mathbf r \left(\frac{(\mathbf v \cdot \mathbf u)}{c^{2}} + 1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}\right)}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right] = \frac{Qq}{r'^{3}}\left[ \frac{[\mathbf v \times [\mathbf u \times \mathbf r]]}{c^{2}\left(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}\right)} + \frac{\mathbf r}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right].   
$$
After that, using $\mathbf r'^{3} = \left(r'^{2}\right)^{\frac{3}{2}} = (r^{2} + \gamma^{2}\frac{(\mathbf r \cdot \mathbf u)^{2}}{c^{2}})^{\frac{3}{2}}$, we can assume, that 
$$
\frac{\mathbf F}{\gamma(1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}})} = \frac{qQ}{(r^{2} + \gamma^{2}\frac{(\mathbf r \cdot \mathbf u)^{2}}{c^{2}})^{\frac{3}{2}}}\!{\left[\frac{[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}]}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}} + \frac{\mathbf r}{1 - \frac{(\mathbf v \cdot \mathbf u)}{c^{2}}}\right]} \Rightarrow
$$
$$
\Rightarrow \mathbf F = \frac{q Q \gamma}{(r^{2} + \gamma^{2}\frac{(\mathbf r \cdot \mathbf u)^{2}}{c^{2}})^{\frac{3}{2}}}\!{\left[\mathbf r + \left[\mathbf v \times \frac{[\mathbf u \times \mathbf r]}{c^{2}}\right]\right]} \qquad (.2).
$$
Using designations
$$
\mathbf E = \frac{Q\gamma \mathbf r}{\left(r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf u \cdot \mathbf r)^{2} \right)^{\frac{3}{2}}}, \quad \mathbf B = \frac{1}{c}[\mathbf u \times \mathbf E],
$$
(.2) can be rewrited as
$$
\mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B].
$$
Of course, you know, that
$$
\mathbf F = \frac{d}{dt}(\frac{m \mathbf v }{\sqrt{1 - \frac{v^{2}}{c^{2}}}}) = \frac{m\mathbf v (\frac{\mathbf v \cdot \mathbf a}{c^{2}})}{\left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}} + \frac{m \mathbf a}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}.
$$
