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 $\theta$.