A general complex electric field When dealing with a plane wave solution to the electric field such as
$$\vec{E}(r,t)=E_{0}\cos(kz-\omega t+\phi)$$
we usually introduce a complex electric field $\tilde{E}(r,t)$ such that $\vec{E}(r,t)=\textrm{Re}\left(\tilde{E}(r,t)\right)$, e.g.
$$\tilde{E}(r,t)=E_{0}e^{i(kz-\omega t+\phi)}$$
I have two questions, both with a mathematical slant.
$1)$ Does there exist an extension of this definition for arbitrary $\vec{E}(r,t)$? One way might be to take the temporal Fourier transform of $\vec{E}(r,t)$ and dismiss all of the positive frequencies. However this might raise issues with causality?
$2)$ There is a huge choice in $\tilde{E}(r,t)$, in particular it seems as if there is an arbitrary choice in $\textrm{Im}\left(\tilde{E}(r,t)\right)$. What additional properties should $\tilde{E}(r,t)$ satisfy so as to define it uniquely?
 A: Your definition in your point 1 is the one that is generally used.
Once you make this decision, there is almost no redundancy in the way you can choose $\mathrm{Im}(\tilde{E})$. Recall that, if two square integrable functions have the same Fourier transform, they are equal almost everywhere (in the measure theoretic sense), i.e. they can only differ on a set of measure nought.
So, your definition in (1) uniquely fixes $\mathrm{Im}(\tilde{E})$, if we require our functions to be continuous.
Now let's answer the causality issue. Yes, in general the imaginary part of $\tilde{E}$ will be acausal, since we have Titmarsh's theorem: 

Given a complex valued function $F:\mathbb{R}\to\mathbb{C}$ on the real line and that $F$ is square-integrable there ($\int_{-\infty}^\infty |F|^2(x)\,\mathrm{d} x<\infty$), the following assertions are all logically equivalent:


*

*$F:\mathbb{R}\to\mathbb{C}$ is the limit of a function $\mathcal{F}:\mathbb{U}\to\mathbb{C}$ holomorphic in the open upper half plane $\mathbb{U}$, i.e. $F(x) = \lim\limits_{y\to0} \mathcal{F}(x+i\,y)$ where $\mathcal{F}$ is also Hardy class $\mathbb{H}^2$, i.e. $f(y) = \int_{-\infty}^\infty |\mathcal{F}(x+i\,y)|^2\mathrm{d}x<\infty,\,\forall y\in\mathbb{R}^+$;

*The real and imaginary parts of $F$ (on the real line) are Hilbert Transforms of one another;

*The Fourier transform $\mathscr{F}(\mathcal{F})(x)$ vanishes for $x<0$.

The Hilbert transform generally maps causal functions into acausal ones, hence your imaginary part will be acausal.
However, this is not a problem because, at the end of any calculation, you restore the negative frequency part by taking $\mathrm{Re}(\tilde{E})$.
A: You are on the right track in thinking about Fourier transforms. Consider an arbitrary scalar function $f(x)$, which we will take to be real; that is
\begin{equation}
f^*(x)=f(x).
\end{equation}
Now we write $f(x)$ in terms of its Fourier transform
\begin{equation}
f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty} dk\; e^{ikx}f(k).
\end{equation}
Complex conjugation yields
\begin{equation}
f^*(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}dk\;e^{-ikx}f^*(k).
\end{equation}
Now we make a change of variables $k\rightarrow -k$:
\begin{equation}
\begin{split}
f^*(x)&=-\frac{1}{2\pi}\int_{\infty}^{-\infty}dk\;e^{ikx}f^*(-k)\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}dk\;e^{ikx}f^*(-k).
\end{split}
\end{equation}
Using the fact that $f(x)$ is real yields the condition
\begin{equation}
f^*(-k)=f(k).
\end{equation}
So, an arbitrary real field (the generalization to vector fields is straightforward) can be written in terms of a complex field (its Fourier transform) subject to the above reality condition.
The field in your example contains only a single frequency in $z$ and $t$, respectively, and so the Fourier transform is trivial and the reality condition is $E_0=E_0^*$.
