In relativistic mechanics the Lorentz force is defined:
$$ \frac{d\mathbf{p}}{dt} = q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right). $$

This looks exactly like the non-relativistic expression.  But while the notation is the same, there are some definitional differences between relativistic and non-relativistic mechanics.

The mistake is that the correct expression for momentum is 
$$
  \mathbf{p} = \gamma m \mathbf{v}, \quad\quad\quad
  \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}, \quad\quad\quad
  v^2 = \mathbf{v}\cdot\mathbf{v} = {v_x}^2 + {v_y}^2 + {v_z}^2
$$

The important thing is that the relativistic Lorentz factor, $\gamma$, depends on the magnitude of velocity, $v$.

Using non-relativistic mechanics when the electric force is applied in the $\hat{y}$ direction, only the $y$-component of velocity changes.  The non-relativistic momentum is just $m\mathbf{v}$, so the $y$-component of velocity changes independently of the $x$- and $z$-components.

Using the correct relativistic formulation, an electric force in the $\hat{y}$ direction affects all three components of velocity.  Changing any one component of momentum will affect the magnitude of velocity, which mixes into the other components of the relativistic momentum.  The $x$-component of velocity can change even if the $x$-component of momentum didn't.

In a reference frame where $\mathbf{B}=0$ and $\mathbf{E}=E_0 \hat{y}$, we have
$$ \frac{d\mathbf{p}}{dt} = q\mathbf{E} $$

To find the final velocity we solve the Lorentz force equation:
$$ \mathbf{p} = \mathbf{p}_0 + q \mathbf{E}\, \Delta t.$$

The $x$-component of momentum doesn't change:
$$ p_x = p_{x,0},$$
but the $x$-component of velocity does!  Any change in $v_y$ means $\gamma \ne \gamma_0$, so $v_x \ne v_{x,0}$.
$$ \gamma m v_x = \gamma_0 m v_{x,0} $$

Looking at both the $x$- and $y$-components:
\begin{align}
  p_{x} &= p_{x,0} & p_{y} &= p_{y,0} + q E_0 \Delta t \\
  \gamma m v \cos\theta &= \gamma_0 m v_0 \cos\theta_0 &
  \gamma m v \sin\theta &= \gamma_0 m v_0 \sin\theta_0 + q E_0 \Delta t.
\end{align}

The final Lorentz factor, $\gamma$, depends on the final velocity, $v$, which makes the algebra a bit of a mess. In principle we can solve the system of two equations for $v$ and $\theta$.

Combining the two equations we can write:
\begin{align}
  \left( \gamma v \right)^2 &= \left( \gamma_0 v_0 \cos\theta_0 \right)^2 + \left( \gamma_0 v_0 \sin\theta_0 + \frac{q}{m} E_0 \Delta t \right)^2 \\
  \frac{v^2}{1-\frac{v^2}{c^2}} &= {\gamma_0}^2 {v_0}^2 + \frac{q^2}{m^2} {E_0}^2 {\Delta t}^2 + 2 \frac{q}{m} E_0 \Delta t\, \gamma_0 v_0 \sin\theta_0
\end{align}