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, $$|\vec E_+|-|\vec E_-|=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0$$
This is an extremely common mistake in introductory EM - from students who actually spend time thinking about the problem, anyway ;-) Use Gauss's law in both cases:
In the case of infinite plates, you do not have the result you give first. A Gaussian cylinder has two disks on either side of the plate, so
$$E_1(2A)=\frac{\sigma A}{\epsilon_0}\rightarrow E_1=\frac{\sigma}{2\epsilon_0}$$
And from superposition you get the total electric field
$$E=\frac{\sigma}{\epsilon_0}$$
You second case is correct, but the charge enclosed by your surface is $Q/2$ relative to the first case (conservation of charge, if you want the same answer you better have the same total charge on the plates), so
$$E_1A=\frac{(\sigma/2) A}{\epsilon_0}\rightarrow E_1=\frac{\sigma}{2\epsilon_0}$$
Which again gets you the same answer when you apply superposition.
Best Answer
I think the best way to answer this question is to actually do the math and physics. From first principles and not some shortcut.
From Couloub's law and the definition of the electric field: $$\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{\mathbf{r}}$$
Consider first an infinite wire of change (we will build the sheet later). For now, we assign a charge density of the entire wire: $\lambda$. Where $\lambda = \frac{dq}{d\ell}$.
The differential form of the electric field equation may then be given as (using the notation from the image):
$$d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} \hat{\mathbf{r}} = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{r^2} d\ell \;\hat{\mathbf{r}}$$
Now, from the image, it should be a bit clear that the electrical field components from the wire in the "up down" ($\hat{\mathbf{y}}$) direction cancel each other out regardless of the value of $R$ and $\ell$. Therefore, let us only consider the electric field in the $\hat{\mathbf{x}}$ direction.
From the geometry, we notice the following:
$$ r = \sqrt{\ell^2 + R^2} = \frac{R}{\cos \theta} $$ $$ d\ell = R d\theta $$ $$ \hat{\mathbf{x}} = \cos \theta \; \hat{\mathbf{r}}$$
Therefore:
$$d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\lambda \cos^2 \theta}{R^2} \frac{R}{\cos \theta} d\theta \;\hat{\mathbf{x}}$$ $$ = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \frac{\cos^2 \theta}{\cos \theta} \;\hat{\mathbf{x}} $$
Now, we want to find the total electric field from the entire length of the wire. Thus, we want to integrate over the entire wire. This means that, intergrating over the angle of $\theta$: $-\frac{\pi}{2} \rightarrow \theta \rightarrow \frac{\pi}{2}$. Therefore:
$$ \vec{E_x} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \frac{\cos^2 \theta}{\cos \theta} \;\hat{\mathbf{x}} $$ $$ = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \;\hat{\mathbf{x}} $$ $$ = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{R} \;\hat{\mathbf{x}}$$
Note that, for an infinite wire, the electric field does depend on your distance from the wire.
However, we want the sheet. We can construct a sheet of chrage by aligning many wires in a row, parallel to each other. Suppose, still using the image, we stack them along the $\hat{\mathbf{z}}$ axis. It should be clear that, like the $\hat{\mathbf{y}}$ component of the electric field cancels itself out when the wire runs along that axis, the sheet also cancels out the contributions from $\hat{\mathbf{z}}$.
We reassign the distance that the point in question is from the sheet as $D$, as $R$ is now between the point and one of the wires (a distance $z$ from the point on the sheet above the point in question) in the entire sheet. This means that $R$ is related now, given by:
$$ R = \sqrt{D^2 + z^2} = \frac{D}{\cos \phi} $$
Where $\phi$ is the angle between the lines $R$ and $D$, similar to how $\theta$ is the angle for the image about (just extrapolate to 3D). Moreover, the surface charge of the sheet is now given by:
$$ \lambda = \sigma dz = \sigma D d\phi $$
$$ \hat{\mathbf{r'}} = \cos \phi \; \hat{\mathbf{x}} $$
Note that the second equation might not make a lot of sense at first; however it is similar to our previous transformation ($ \hat{\mathbf{x}} = \cos \theta \; \hat{\mathbf{r}}$) execpt that the direction is a new offset from $\hat{\mathbf{r'}}$.
If we take the answer for the electric field via a line of charge and put it into a differential form:
$$ d\vec{E_{r'}} = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{R} \;\hat{\mathbf{r'}} $$
Subsituting:
$$ d\vec{E_x} = \frac{1}{4\pi\epsilon_0} \frac{2 \sigma D }{D} \frac{\cos \phi}{\cos \phi} d\phi \;\hat{\mathbf{x}} $$ $$ = \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) d\phi \;\hat{\mathbf{x}} $$
Finally, again, as with the wire, we integrate over the entire sheet: $-\frac{\pi}{2} \rightarrow \phi \rightarrow \frac{\pi}{2}$
$$ \vec{E_x} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) d\phi \;\hat{\mathbf{x}} $$ $$ = \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\phi \;\hat{\mathbf{x}} $$ $$ = \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) \left( \pi \right) = \frac{\sigma}{2 \epsilon_0} $$
As you can see, because of the geometry of the infinite sheet, the dependence on the distance from the sheet fell out of the equation (with no approximations, for the most part). Doing the calculation from first principles, we have obtained an equation for the electric field via an infitie plate that one would normally find a textbook.
If your question asked for the actual reason (and not how we know it), this entire derivation is a consequence from Coulomb's law. To ask why Coulomb's law is as it is, is outside the scope of this answer (and physics?).