Does substituting the Lorentz-transformed scalar and vector potentials into the $E$-field formula yield the correct Lorentz-transformed $E$-field? If we switch from one inertial frame to a different inertial frame with a relative velocity of $\mathbf{v}$, we could transform the scalar and vector potentials thusly:
$$\varphi' = \gamma \left( \varphi - \mathbf{A}\cdot \mathbf{v} \right) $$
$$\mathbf{A}' = \mathbf{A} - \frac{\gamma \varphi}{c^2}\mathbf{v} + \left(\gamma - 1\right) \left(\mathbf{A}\cdot\mathbf{\hat{v}}\right) \mathbf{\hat{v}}$$
Source: The Cambridge Handbook of Physics Formulas, G. Woan, Cambridge University Press, 2010, ISBN 978-0-521-57507-2.
It seems logical that the expression of the electric field in terms of the potentials would be unchanged before and after the Lorentz transformation:
$\mathbf{E} = -\mathbf{\nabla} \varphi - \frac{\partial \mathbf{A}}{\partial t}$
$\mathbf{E'} = -\mathbf{\nabla} \varphi' - \frac{\partial \mathbf{A'}}{\partial t'}$
Consider a straight line parallel to $\mathbf{\hat{z}}$ along which the scalar potential $\varphi$ and vector potential $\mathbf{A}$ both remain spatially-uniform and changing with time.
Let's have an independent inertial observer that moves with relative velocity $\mathbf{v}$ in a direction parallel to $\mathbf{\hat{z}}$
In this case, we can confirm rather easily on this straight line that minus the spatial-derivative of the electric scalar potential $-\nabla\varphi'$ is zero along $\mathbf{\hat{z}}$ both before and after the transformation (i.e. $\frac{\partial \varphi}{\partial z} = \frac{\partial \varphi'}{\partial z'} = 0$). On the other hand, can we say the same thing about minus the time-derivative of the magnetic vector potential $-\frac{\partial \mathbf{A'}}{\partial t'}$?
As a result of a Lorentz boost of velocity $\mathbf{v}$, the vector potential transforms by the amount:
$$\mathbf{A}' - \mathbf{A} = - \frac{\gamma \varphi}{c^2}\mathbf{v} + \left(\gamma - 1\right) \left(\mathbf{A}\cdot\mathbf{\hat{v}}\right) \mathbf{\hat{v}}$$
On the right-hand side of this equation, our only variables are $\varphi$ and $\mathbf{A}$. In contrast, $\gamma$, $\mathbf{v}$, and $\mathbf{\hat{v}}$ are all properties of the Lorentz boost and are therefore constant, as is $c^2$.
[Edit: Let's not concern ourselves with taking the derivative of this equation with respect to time. Let's consider the derivative of this equation with respect to changes of $\varphi$ or $\mathbf{A}$.]
Expansion of the first term on the R.H.S. reveals a leading term proportional to $\mathbf{v}$ and parallel to $\mathbf{v}$.
Expansion of the second term on the R.H.S. reveals a leading term proportional to $\mathbf{v^2}$ and parallel to $\mathbf{v}$.
This means that the transformation of the magnetic vector potential is such that it differs by a vector equal to $\mathbf{A}' - \mathbf{A}$ that is parallel to the Lorentz boost. Consequently, the derivative of $\mathbf{A}' - \mathbf{A}$ with respect to $\varphi$ or $\mathbf{A}$ is also parallel to the Lorentz boost. Since $\varphi$ and $\mathbf{A}$ may not be constant, this difference implies the existence of a contribution to the component of the electric field parallel to the Lorentz boost depending on our time-varying potentials $\varphi$ and $\mathbf{A}$.
The problem I see here is that if we add a zero-valued contribution to the electric field
$-\mathbf{\nabla_z} \varphi' - (- \mathbf{\nabla_z} \varphi) = 0$
with a non-zero contribution to the electric field
$- \frac{\partial \mathbf{A_z}'}{\partial t'} - (- \frac{\partial \mathbf{A_z}}{\partial t}) \neq 0$
in the direction of the Lorentz boost (parallel to $\mathbf{\hat{z}}$), the sum would be non-zero. How then could these transformations of the electromagnetic potentials [Edit: Clarification - …using the two formulas from "The Cambridge Handbook of Physics Formulas" presented at the top of this post…] be consistent with the fact that
 $\mathbf{E_\parallel}' = \mathbf{E_\parallel}$?
 A: It is easier to decompose each vector into parallel and perpendicular components, since the Lorentz transformations leave the perpendicular components unchanged.
Let us use the convention that bold-faced symbols are standard three component vectors. Denoting $\boldsymbol{\beta} = \mathbf v/c$ and a general four-vector by $f$ we shall use the following four-vector transformations:
\begin{align}
\tag{1} \nabla &=\left(\frac{\partial}{\partial(ict)},\boldsymbol{\nabla}\right)\\[5pt]
\tag{2} A & =\left(\frac{i \phi}{c},\mathbf{A} \right) \\[5pt]
\tag{3} \mathbf{f'_{\parallel}} &= \gamma(\mathbf{f_{\parallel}}+i \boldsymbol{\beta} f_0)\\[5pt]
\tag{4} f'_0 &= \gamma(f_0 - i \boldsymbol{\beta} \cdot \mathbf{f_{\parallel}})
\end{align}
Now we shall use the definition 
$$\mathbf{E'_{\parallel}} = - \nabla'_{\parallel} \phi' - \dfrac{\partial \mathbf{A'_{\parallel}}}{\partial t'}$$
and substitute eqn. $(1) - (4)$ in the above,
\begin{align}
\tag{5}
\mathbf{E'_{\parallel}} = -\gamma\left(\nabla_{\parallel}+\frac{\mathbf v}{c^2}\frac{\partial}{\partial t}\right)\gamma(\phi - \mathbf{v \cdot A}) - \gamma \left(\frac{\partial}{\partial t}+\mathbf v \cdot \nabla_{\parallel}\right)\gamma\left(\mathbf{A_{\parallel}}-\frac{\mathbf v}{c^2}\phi\right)
\end{align}
The right side of eqn. $(5)$ when simplified generates eight terms, out of which two cancel each other because of $\pm (\mathbf{v}/c^2) \partial{\phi}/\partial{t}$. Also, the term with $(\mathbf{v}\cdot\nabla)\mathbf{A}_{\parallel}$ cancels with $-\nabla_{\parallel}(\mathbf{v}\cdot\mathbf{A})$ since $\mathbf{v}$ is a constant. In the end only four terms remain.
I will let you figure out that those terms can be reduced to
$$
\tag{6}
\mathbf{E'_{\parallel}} = -\gamma^2 \left(\nabla_{\parallel} \phi - \dfrac{\partial \mathbf{A_{\parallel}}}{\partial t} \right)(1 - v^2/c^2).
$$
Note that we can use $\gamma^2 (1 - \beta^2) = 1$ in $(6)$ yielding us the desired result,
$$
\tag{7}
\mathbf{E'_{\parallel}} = -\nabla_{\parallel} \phi - \dfrac{\partial \mathbf{A_{\parallel}}}{\partial t} = \mathbf{E_{\parallel}}.
$$
EDIT:
In fact the transformations used here are same equations OP cited from The Cambridge Handbook of Physics Formulas.
Proof:
From eqns. $(2)$ and $(4)$,
\begin{align}
\frac{i\phi'}{c} &= \gamma\left(\frac{i\phi}{c} - i \frac{\mathbf{v}}{c}\cdot \mathbf{A}\right) \\[5pt]
\phi' &= \gamma\left(\phi - \mathbf{v}\cdot\mathbf{A}\right)
\end{align}
Also, from eqns. $(2), (3)$ and $(5)$,
\begin{align}
\mathbf{A'} &= \mathbf{A'_{\perp} + \mathbf{A'_{\parallel}}}\\[5pt]
   &= \mathbf{A_{\perp}} + \gamma \left(\mathbf{A_{\parallel}} + i \frac{\mathbf{v}}{c} \frac{i \phi}{c}\right)\\[5pt]
&= \left(\mathbf{A_{\perp}} + \mathbf{A_{\parallel}}\right) + (\gamma-1)\mathbf{A_{\parallel}} - \frac{\gamma \phi}{c^2}\mathbf{v}\\[5pt]
&= \mathbf{A} - \frac{\gamma \phi}{c^2} \mathbf{v} + (\gamma-1)(\mathbf{A}\cdot\mathbf{\hat{v}})\mathbf{\hat{v}}
\end{align}
where the last equality comes from resolving $\mathbf{A}$ into parallel component along $\mathbf{\hat v}$.
