The first thing to remember is that the Fourier transform $\mathcal{F}$ is linear: $$
\mathcal{F}(\alpha f+\beta g)=\alpha \mathcal{F}(f)+\beta \mathcal{F}(g)
$$
It also changes differentiation into multiplication. Let $F(\omega)$ be the transform of $f(t)$. Then:
$$
\mathcal{F}(D_tf)(\omega)=\int_{-\infty}^{\infty}D_tf(t)e^{-j\omega t}\mathrm{d}t = f(t)e^{-j\omega t}|_{-\infty}^{\infty}+j\omega\int_{-\infty}^{\infty} f(t)e^{-j\omega t}\mathrm{d}t = 0 + j\omega F(\omega)
$$
For this to actually be correct, the boundary term from integration by parts has to disappear in the limit. This imposes some technical conditions. Also, if $f$ happens to be periodic, the Fourier transform (likely) won't exist in the standard sense and may require the use of distributions. I won't dwell on these issues here (mostly because I don't actually know a whole lot about all the theory behind these concerns), but perhaps you can make a separate question specifically about when this process works, or maybe someone else can add another answer regarding this.
Anyway, I'll just do the formal manipulations here (as is usually done in physics classes). In your case, we have functions of several variables $({\bf r}, t)$. However, this isn't a problem. For example:
$$
\mathcal{F}(\partial_t{\bf e})({\bf r},\omega)=\int_{-\infty}^{\infty}\partial_t{\bf e}({\bf r},t)e^{-j\omega t}\mathrm{d}t= j\omega {\bf E}({\bf r},\omega)
$$
Here, the process is exactly the same as the above since we can keep ${\bf r}$ fixed and so you can use integration by parts on $\partial_t$. If you're worried about vector terms, notice that this is no different from integrating any other function ${\bf f}(u)$ with values in $\mathbb{R^3}$ (I'll omit the integration limits because they don't really matter):
$$
\int{\bf f}(u)\mathrm{d}u=\left(\int f_x(u)\mathrm{d}u,\int f_y(u)\mathrm{d}u,\int f_z(u)\mathrm{d}u\right)
$$
Or, in component notation ($i=x,y,z$):$$
\left(\int{\bf f}(u)\mathrm{d}u\right)_i=\int f_i(u)\mathrm{d}u$$
The curl is also not problematic as the spatial derivatives commmute with time integration:
$$
\int\partial_x{\bf f}({\bf r},t)\mathrm{d}t=\partial_x\int{\bf f}({\bf r},t)\mathrm{d}t
$$
Because of this, you can immediately conclude that $\mathcal{F}(\partial_x{\bf f})=\partial_x\mathcal{F}({\bf f})$. Writing out only the $x$-component:
$$
\left(\mathcal{F}(\nabla\times{\bf f})\right)_x=\mathcal{F}(\partial_yf_z-\partial_zf_y)=\partial_yF_z-\partial_zF_y = (\nabla\times{\bf F})_x
$$
Here ${\bf F}=\mathcal{F}({\bf f})$. The other two components work in exactly the same way. Finally, take the equation $$\nabla\times{\bf h}={\bf j}+\epsilon_0\partial_t{\bf e}+\partial_t{\bf p}$$
and apply $\mathcal{F}$ to both sides. By linearity and what we've said about $\nabla\times$, we find
$$
\mathcal{F}(\nabla\times{\bf h})=\mathcal{F}({\bf j}+\epsilon_0\partial_t{\bf e}+\partial_t{\bf p})$$
$$
\nabla\times\mathcal{F}({\bf h})=\mathcal{F}({\bf j})+\epsilon_0\mathcal{F}(\partial_t{\bf e})+\mathcal{F}(\partial_t{\bf p})
$$
Keeping in mind what $\mathcal{F}$ does to $\partial_t$, we finally get:
$$
\nabla\times{\bf H}={\bf J}+j\omega\epsilon_0{\bf E}+j\omega{\bf P}
$$
The other equation follows suit.
As one commenter pointed out, this is a Fourier transform on time. It is also possible to do a Fourier transform on the spatial coordinates. These two approaches are complementary; often, both Fourier transforms are taken and so we change from $({\bf r},t)$-space to $({\bf k},\omega)$-space where ${\bf k}$ is the wave vector.
Let's formulate the definition of curl slightly more precisely in the form of a definition/theorem. I'll also not use boldface objects, simply for ease of typing
Definition/Theorem.
Let $A\subset \Bbb{R}^3$ be open, $F: A \to \Bbb{R}^3$ be $C^1$. Then, there is a unique continuous function $H:A \to \Bbb{R}^3$, such that for every $p\in A$, for every $\epsilon > 0$, for every "nice" surface $S\subset A$, there is an open neighbourhood $U$ of $p$ in $S$ such that for every "nice" oriented surface $M$ with $p\in M\subset U$, with outward normal vector field $n(\cdot)$ (which is simply the restriction of the outward normal of $S$ to $M$), boundary $\partial M$ and tangent vector field $\tau(\cdot)$ on $\partial M$, we have:
\begin{align}
\left|\dfrac{1}{|M|}\int_{\partial M}\langle F, \tau\rangle\, dl - \langle H(p), n(p)\rangle\right| < \epsilon\tag{1}
\end{align}
(this is the more precise meaning of the limit you're talking about)
In this case, because $H$ is unique, we can give it the name $\text{curl}(F)$. In fact, we can show that
\begin{align}
H =
\left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \boldsymbol{\hat k} \tag{2}
\end{align}
In the above paragraph, "nice oriented surface" means nice enough so that Stoke's theorem can be applied; for example a smooth two-dimensional, oriented manifold-with-boundary (or however much you wanna weaken the hypothesis... because each book presents it with varying levels of generality... just as long as Stoke's theorem can be applied).
Note that for the above definition of curl to make sense, we have to first show the existence and uniqueness of such a vector field $H$. We shall start by showing the uniqueness of $H$. So, we assume such a $H$ exists, and then prove its components are determined according to the formula $(2)$; this will complete the proof of uniqueness.
Proof of Uniqueness
I'll carry out the computation in detail for proving $H_x = \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z}$, and leave the other two to you (it's simply a matter of renaming $x,y,z$). We prove this equality pointwise of course. So, fix a point $p \in A$; then for any $\delta > 0$ such that the closed cube $C_{p,\delta} = p + [-\delta,\delta]^3$ (which is the closed cube centered at $p$ of sidelength $2\delta$) lies entirely inside $A$ (note that since $A$ is open, there are infinitely many such $\delta>0$), we define $M^{\delta} := \{p_1\}\times [p_2-\delta, p_2 + \delta]\times [p_3-\delta, p_3 + \delta]$. This is a piece of a plane which we shall orient so that it has a constant outward normal vector field $n = e_1 \equiv \boldsymbol{i}$. Now, we calculate: note that $\partial M^{\delta}$ has $4$-pieces, and the unit tangent vector along these boundary paths is constant, so (if you're very careful with signs... which I hope I didn't make any sign mistakes), we get
\begin{align}
\dfrac{1}{|M^{\delta}|} \int_{\partial M^{\delta}}\langle F, \tau \rangle\, dl &= \dfrac{1}{4\delta^2}
\bigg[\int_{-\delta}^{\delta} F_2(p_1, p_2 + y, p_3-\delta) - F_2(p_1, p_2 + y, p_3+\delta) \, dy\bigg]\\
&+\dfrac{1}{4\delta^2}\bigg[ \int_{-\delta}^{\delta} F_3(p_1, p_2+\delta, p_3+z) - F_3(p_1, p_2-\delta, p_3+z)\, dz\bigg]
\end{align}
For each term, we apply the mean-value theorem for integrals (which we can use since everything is continuous), to obtain some $\eta \in [p_2-\delta, p_2+\delta]$ and some $\zeta\in [p_3-\delta, p_3+\delta]$ such that
\begin{align}
\dfrac{1}{|M^{\delta}|} \int_{\partial M^{\delta}}\langle F, \tau \rangle\, dl &=
\dfrac{1}{4\delta^2}\bigg[2\delta \cdot F_2(p_1, \eta, p_3-\delta) - 2\delta \cdot F_2(p_1, \eta, p_3+\delta)\bigg] \\
&+ \dfrac{1}{4\delta^2}\bigg[2\delta \cdot F_3(p_1, p_2+\delta, \zeta) - 2\delta \cdot F_3(p_1, p_2-\delta, \zeta)\bigg] \\\\
&=\dfrac{F_3(p_1, p_2+\delta, \zeta) - F_3(p_1, p_2-\delta, \zeta)}{2\delta} -\dfrac{F_2(p_1, \eta, p_3+\delta) - F_2(p_1, \eta, p_3-\delta)}{2\delta} \\
&= \dfrac{\partial F_3}{\partial y}(p_1, \alpha, \zeta) - \dfrac{\partial F_2}{\partial z}(p_1, \eta, \beta),
\end{align}
for some $\alpha\in [p_2-\delta, p_2+\delta], \beta\in [p_3-\delta, p_3+\delta]$, using the mean-value theorem for derivatives (which can certainly be applied since we assumed $F$ is $C^1$).
Quick summary: what we showed so far is that for every $p\in A$ and every $\delta>0$ such that the cube $C_{p,\delta}$ lies inside $A$, if we define $M^{\delta}$ as above to be the plane centered at $p$ with normal pointing in $e_1$ direction, then, there exist points $a_{p,\delta},b_{p,\delta} \in M^{\delta} \subset C_{p,\delta}$ inside the surface (in particular inside the cube) such that
\begin{align}
\dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle \, dl &=
\dfrac{\partial F_3}{\partial y}(a_{p,\delta}) - \dfrac{\partial F_2}{\partial z}(b_{p,\delta})
\end{align}
From here, it is a simple matter of using the continuity of partial derivatives. Here's the full $\epsilon,\delta$ argument to finish it off: let $p\in A$ and $\epsilon> 0$ be arbitrary. By our hypothesis of $(1)$, there is an open $U$ such that blablabla. Now, for this given $\epsilon > 0$, let's choose $\delta > 0$ small enough so that
- the cube $C_{p,\delta}$ lies inside $U$
- the $\delta$ "works" for the continuity of $\dfrac{\partial F_3}{\partial y}$ and $\dfrac{\partial F_2}{\partial z}$ at the point $p$
(so really we have to take a minimum of several $\delta$'s). Then, we choose the oriented plane $M^{\delta}$ as defined above (this plane lies inside $U$ by construction, because of how small $\delta$ is).
\begin{align}
\left|\left(\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_2}{\partial z}(p)\right) - H_1(p)\right| &= \left|\left(\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_2}{\partial z}(p)\right) - \langle H(p), n(p)\rangle\right| \\\\
&\leq \left|\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_3}{\partial y}(a_{p,\delta})\right|
+ \left|-\dfrac{\partial F_2}{\partial z}(p) + \dfrac{\partial F_2}{\partial z}(b_{p,\delta})\right|\\
&+ \left|\left(\dfrac{\partial F_3}{\partial y}(a_{p,\delta}) - \dfrac{\partial F_2}{\partial z}(b_{p,\delta})\right) - \dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle\, dl \right| \\
&+ \left|\dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle\, dl - \langle H(p), n(p) \rangle \right| \\\\
&\leq 4\epsilon
\end{align}
(each absolute value is $\leq \epsilon$ based on everything I've said above, and by the choice of $\delta$). Since the point $p$ and $\epsilon > 0$ are arbitrary, the inequality above shows that
\begin{align}
H_1 &= \dfrac{\partial F_3}{\partial y} - \dfrac{\partial F_2}{\partial z}
\end{align}
As a recap of the proof idea: choose a small plane $M^{\delta}$ with outward normal pointing along $e_1$; it is the flatness of the plane (which is inherently adapted to cartesian coordinates), along with the ease of boundary parametrization which makes the resulting line integral easy to calculate. Then, simply calculate everything, and use the mean-value theorems for derivatives and integrals (this is one way to fill in the gaps for the arugments you typically see in physics texts, which say "let's keep things up to first order only" and where they use $\approx$ everywhere); finally we complete it off with a standard $\epsilon,\delta$ continuity argument.
Some remarks is that for this argument to work I've had to assume $F$ is $C^1$, so that I can apply the mean-value theorems twice, and finally finish it off with a continuity argument. I'm not sure if this proof can be strengthened so that we only have to assume $F$ is differentiable (rather than $C^1$).
Proof of Converse
Now we show the existence of such a vector field $H$; for this we'll show that $\text{curl}F$, defined by equation $(2)$ satisfies the conditions of $(1)$. Like I mentioned in the comments, I'm not sure how to do this without already appealing to Stokes theorem. With Stokes' theorem, this becomes quite simple.
Let $p\in A$, $\epsilon > 0$ and let $S\subset A$ be any "nice surface". Since $\langle\text{curl}(F), n\rangle$ is a continuous function on $S$, there is an open neighbourhood $U$ around $p$ in $S$ such that for all $q\in U$,
\begin{align}
\left|\langle \text{curl}F(q), n(q)\rangle - \langle \text{curl}F(p), n(p)\rangle\right| & \leq \epsilon
\end{align}
Now, for any "nice surface" $M\subset U$ (with unit normal being the restriction of the one already on $S$), we have by Stokes theorem:
\begin{align}
\left|\dfrac{1}{|M|}\int_{\partial M}\langle F,\tau\rangle \, dl - \langle \text{curl} F(p), n(p) \rangle\right|
&= \dfrac{1}{|M|}\left|\int_M \langle \text{curl}F, n\rangle \, dA - \int_M \langle \text{curl} F(p), n(p) \rangle\, dA\right|
\\
&\leq \dfrac{1}{|M|}\int_M \left|\langle\text{curl }F, n\rangle -
\langle\text{curl }F(p), n(p)\rangle \right| \, dA \\
& \leq \dfrac{1}{|M|} \epsilon |M| \\
&= \epsilon.
\end{align}
This completes the proof of existence.
Best Answer
We know that the only component of $\mathbf{E}$ is along the $\mathbf{e_{x}}$
Therefore $$ \nabla \times \mathbf{E} = \partial_zE_x \mathbf{e_{y}} - \partial_yE_x \mathbf{e_{z}} = \partial_zE_x \mathbf{e_{y}} $$ We also have $E_x = f(z)$ Thus the only component for $\mathbf{H}$ is along the $\mathbf{e_{y}}$ or $H_y$
We have $$ \nabla \times \mathbf{E} = -j\omega \mu \mathbf{H} $$ Which given $$ \mathbf{E} = E_x(z)\mathbf{e_{x}} $$ we have $$ \frac{\partial E_x}{\partial z}\mathbf{e_{y}} = - j \omega \mu H_y\mathbf{e_{y}} $$
We have $$ H_y = \frac{-1}{j\omega \mu}\frac{\partial E_x}{\partial z} = \frac{j}{\omega \mu}\frac{\partial E_x}{\partial z} $$ If we apply $$ E_x = E^{+}\mathrm{e}^{-jkz} + E^{-}\mathrm{e}^{jkz}\\ \partial_z E_x = -jk E^{+}\mathrm{e}^{-jkz} + jkE^{-}\mathrm{e}^{jkz} = - jk \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right] $$ we then have $$ H_y =- \frac{j}{\omega \mu}\cdot jk \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right]\\ H_y =\frac{k}{\omega \mu} \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right]\\ H_y =\frac{1}{\eta} \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right] $$