I am really confused with expressions under Lorentz transformation. I will try to showcase my confusion via an easy and very popular example:

If we have two inertial systems $S$ and $S'$, and $S'$ is moving relative to S with velocity $v$.
A point charge is in the origin of $S'$. Initially at $t=t'=0$ their origins are at the same point.

In $S'$:

We only have electric field: $\vec E'= \frac{1}{4\pi \epsilon\_0}\frac{\vec r'}{|\vec r'|^3}$

There is no magnetic field: $\vec B'=0$

We want to know the electric field and magnetic field at point $P=(0,b,0)$.

This point is in $S$ system as $\vec P=(0,b,0)$.

This point is in $S'$ system as $\vec P'=(x',b,0)$. Performing Lorentz transformation of the kind $x'=\gamma(x -vt)$. We know that $x=0$, therefore $x'=-v\gamma t$. Then $\vec P'=(-v\gamma t,b,0)$. I consider that these are the coordinates of point $\vec P'$ observed by someone in the system $S'$, but via the coordinates of the same point $\vec P$ observed by someone in the rest frame.

But we can also express the coordinates of $\vec P'=(-vt',b,0)$. Here the coordinates are directly related to the frame at motion, and no Lorentz transformation takes place.

Anyway, we perform Lorentz transformations etc and for the x component of the Electric field we have:

$$E\_x=E\_x'=\frac{1}{4\pi \epsilon\_0} \frac{-vt'}{(b^2 + v^2t'^2)^{\frac 32 }}=\frac{1}{4\pi \epsilon\_0} \frac{-v\gamma t}{(b^2 + \gamma^2v^2t^2)^{\frac 32 }}$$

Both of the above expressions are related to $E\_x'$.

When the observer is in the rest frame of the particle and no LT takes place, the expression for the electric field at $\vec P'$ is:

$$E\_x'=\frac{1}{4\pi \epsilon\_0} \frac{-vt'}{(b^2 + v^2t'^2)^{\frac 32 }}$$

Now if we want to express the x-component of the electric field, but by utilizing the coordinates from the rest frame, we would get:

$$E\_x'=\frac{1}{4\pi \epsilon\_0} \frac{-v\gamma t}{(b^2 + \gamma^2v^2t^2)^{\frac 32 }}$$

But if I want to know the expression of the x-component, described by an observer at the rest frame $S$, while taking into consideration the coordinates from the frame at motion $S'$, we need to perform the following LT: $x=\gamma(x'+vt)$ and because $x'=-vt$, I would get $x=0$, which is correct because in the rest frame the point where we want to find the electric field is $\vec P(0,b,0)$ which has a $0$ for the x component. As a result $E\_x=0$. But this is not possible since $E\_x=E\_x'$ for this case.

So what exactly I am not understanding here?