Coulomb's law is indeed a special case between two point charges. to find the force between a point charge and a plate, you would have to integrate the equation over the plate surface to calculate the contributions from all infinitesimal charge elements.
It's more practical to figure out what the electric field is, and then use $\vec{F}_e = q\vec{E}$ to find if there is a force applied to your test charge.
The electric field from an infinite plate with uniform surface charge density is given by $\vec{E} = \frac{\sigma}{2\epsilon_0}\vec{a}_n$ where the $\vec{a}_n$ vector is pointing away from the plate. Therefore, if you have two parallel plates (sufficiently large compared to the distance between them to be considered infinite) with the same $\sigma$ charge density, the electric field between the two will be null, and no force will be exerted on the test charge.
The mistakes you are doing are:
You didn't considered the flux coming from them in between them. You have to take all the flux in all directions coming from them. You should take the gaussian across the surface of the plane otherwise you will get wrong result.
$|\vec E_+|=|\vec E_-|=\frac{\sigma}{2\epsilon_0}$ and not $\frac{-\sigma}{2\epsilon_0}$ for $|\vec E_-|.\space$ $\sigma$ is the magnitude of the charge density.
You are incorrectly adding the fields which gave you $0$ inside. The magnitudes have to be added when directions are same and subtracted when directions are opposite.
This is what we get from Gauss's law: $$\vec{E}=\frac{\sigma}{2\epsilon_0}\hat r$$
where, $$|\vec{E}|=\frac{\sigma}{2\epsilon_0}$$where $\sigma$ is the magnitude of surface charge density
So, outside, if direction of $\vec{E_+}$ is $\hat r$ then, direction of $\vec{E_-}$ is $-\hat r$ $$\vec{E_+}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_-}=\frac{\sigma}{2\epsilon_0}(-\hat r)$$$$\vec{E_+}+\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r+\frac{-\sigma}{2\epsilon_0}\hat r$$ $$=0$$
Inside, both $\vec E_+$ and $\vec E_-$ has same direction $\hat r$ $$\vec{E_+}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_+}+\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r+\frac{\sigma}{2\epsilon_0}\hat r=\frac{\sigma}{\epsilon_0}\hat r$$
Talking in magnitudes, inside, the magnitudes have to be added, $$|\vec E_+|+|\vec E_-|=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$$
outside, they have to be subtracted, $$|\vec E_+|-|\vec E_-|=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0$$
Best Answer
Hint :
Suppose that a plane $\;\mathcal P_1\;$ parallel to the $\;xy\;$ plane at $\;z=0\;$ has a uniform surface charge density $\;\sigma_1$. Then
\begin{equation} \mathbf{E}_1\left(x,y,z\right)= \left. \begin{cases} \boldsymbol{+}\dfrac{\sigma_1}{2\epsilon_0}\mathbf{k},\, \text{for } z>0 \vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \boldsymbol{-}\dfrac{\sigma_1}{2\epsilon_0}\mathbf{k},\, \text{for } z<0 \end{cases}\right\} \tag{01}\label{eq01} \end{equation} where $\;\mathbf{k}\;$ the unit vector on axis $\;z$.
Now, take an other plane $\;\mathcal P_2\;$ parallel to the $\;xy\;$ plane at $\;z=h\;$ with a uniform surface charge density $\;\sigma_2\;$ producing its own $\;\mathbf{E}_2\;$ and find $\;\mathbf{E}\left(x,y,z\right)=\mathbf{E}_1\left(x,y,z\right)+\mathbf{E}_2\left(x,y,z\right)$.
Proof of equation \eqref{eq01} using Coulomb's Law only
In Figure-01 a test charge $\;q\;$ is at a height $\;h\;$ over a plane $\;\mathcal P\;$ with uniform surface charge density $\;\sigma$. Let a ring of finite radius $\;R\;$ and of infinitesimal width $\;\mathrm dR$. If we take a piece of the ring of infinitesimal length $\;\delta\ell\;$ then the infinitesimal area $\;\mathrm dR\cdot\delta\ell\;$ carries charge $\;\sigma\cdot\mathrm dR\cdot\delta\ell\;$ and exerts on the charge $\;q\;$ a force
\begin{equation} \delta\mathbf{f}=\dfrac{q\cdot\left(\sigma\cdot\mathrm dR\cdot\delta\ell\right)}{4\pi\epsilon_{0}r^2}\dfrac{\mathbf{r}}{\;r\;} \tag{02}\label{eq02} \end{equation} This force has components normal and parallel to the plane $\;\mathcal P\;$ \begin{equation} \delta\mathbf{f}=\delta \mathbf{f}_{\boldsymbol z}\boldsymbol{+}\delta \mathbf{f}_{\boldsymbol{\rho}} \tag{03}\label{eq03} \end{equation} If we integrate with respect to $\;\ell\;$ we have \begin{equation} \underbrace{\oint_\ell\delta\mathbf{f}}_{\mathrm d \bf F}=\underbrace{\oint_\ell\delta \mathbf{f}_{\boldsymbol z}}_{\mathrm d \mathbf{F}_{\!\boldsymbol z}}\boldsymbol{+}\underbrace{\oint_\ell\delta \bf f_{\boldsymbol{\rho}}}_{\mathrm d \bf F_{\! \boldsymbol{ \rho}}} \tag{04}\label{eq04} \end{equation} or \begin{equation} \mathrm d \mathbf{F}=\mathrm d \mathbf{F}_{\!\boldsymbol z}\boldsymbol{+}\mathrm d \mathbf{F}_{\! \boldsymbol{ \rho}} \tag{05}\label{eq05} \end{equation} Because of the rotational symmetry with respect to the vertical $\;z-$axis the components $\;\delta \bf f_{\boldsymbol{\rho}}\;$ cancel out so $\;\mathrm d \bf F_{\!\boldsymbol{ \rho}}=\boldsymbol{0}\;$ and \begin{equation} \mathrm d \mathbf{F}=\mathrm d \mathbf{F}_{\!\boldsymbol z}=\dfrac{q\sigma}{2\epsilon_{0}}\dfrac{R\cos\theta\mathrm dR}{r^2}\mathbf{k} \tag{06}\label{eq06} \end{equation} Now from the geometry of this configuration \begin{align} R &=h\tan\theta \tag{07.1}\label{eq07.1}\\ \mathrm dR &=\dfrac{h}{\cos^2\theta}\mathrm d\theta \tag{07.2}\label{eq07.2}\\ r^2 &=\dfrac{h^2}{\cos^2\theta} \tag{07.3}\label{eq07.3} \end{align} and replacing in \eqref{eq06} \begin{equation} \mathrm d \mathbf{F}=\dfrac{q\sigma}{2\epsilon_{0}}\sin\theta\,\mathrm d\theta\,\mathbf{k} \tag{08}\label{eq08} \end{equation} Integrating \begin{equation} \mathbf{F}=\dfrac{q\sigma}{2\epsilon_{0}}\left(\int\limits_{0}^{\pi/2}\sin\theta\,\mathrm d\theta\right)\mathbf{k}=\dfrac{q\sigma}{2\epsilon_{0}}\Bigl[-\cos\theta\Bigr]_{0}^{\pi/2}\mathbf{k}=\dfrac{q\sigma}{2\epsilon_{0}}\mathbf{k} \tag{09}\label{eq09} \end{equation} so \begin{equation} \mathbf{E}=\dfrac{\mathbf{F}}{q}= \dfrac{\sigma}{2\epsilon_{0}}\,\mathbf{k} \tag{10}\label{eq10} \end{equation} a result independent of the position coordinates of the test charge $\;q$.
In Figure-02 we see the force exerted on $\;q\;$ by a ring of infinitesimal width, equation \eqref{eq08}.