Pre-envelope of ${\Pi}_{a}(t)\cos(2{\pi}f_{0}t)$
The pre-envelope is also called analytic signal. Its Fourier transform is indeed given by the expression in your question:

$$X_+(f)=2X(f)u(f)\tag{1}$$

where $X(f)$ is the Fourier transform of the original signal, and $u(f)$ is the unit step function. Obviously, $X_+(f)$ has only positive frequency components. The analytic signal $x_+(t)$ with its Fourier transform given by $(1)$ is necessarily a complex-valued signal:

$$x_+(t)=x(t)+j\mathcal{H}\{x(t)\}\tag{2}$$

where $\mathcal{H}\{\cdot\}$ denotes the Hilbert transform. Note that the analytic signal is not the same as the complex envelope. For a band pass signal $x(t)$, the complex envelope is a low pass signal, whereas the analytic signal is a band pass signal.

From $(2)$, in order to compute the analytic signal you need to compute the Hilbert transform of $x(t)$:

$$\hat{x}(t)=\mathcal{H}\{x(t)\}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau\tag{3}$$

For the given signal you get

$$\hat{x}(t)=\frac{1}{\pi}\int_{-a}^a\frac{\cos(2\pi f_0\tau)}{t-\tau}d\tau\tag{4}$$

The result of $(4)$ can be written in terms of the cosine integral $\text{Ci}(x)$ and the sine integral $\text{Si}(x)$:

$$\hat{x}(t)=\frac{1}{\pi}\left\{\cos(2\pi f_0t)\left[\text{Ci}(2\pi f_0(t+a))-\text{Ci}(2\pi f_0(t-a))\right]+\\\sin(2\pi f_0)\left[\text{Si}(2\pi f_0(t+a))-\text{Si}(2\pi f_0(t-a))\right]\right\}\tag{5}$$

I don't think that anybody expected you to come up with that awful expression. Anyway, for large values of $f_0$, the first term in $(5)$ becomes very small, and the second term converges to $\sin(2\pi f_0)\Pi_a(t)$. So for large $f_0$ you get the expected result

$$x_+(t)\approx x(t)+j\sin(2\pi f_0)\Pi_a(t)=\Pi_a(t)e^{j2\pi f_0 t}\tag{6}$$

Obviously, the respresentation

$$x(t)=\text{Re}\left\{\Pi_a(t)e^{j2\pi f_0 t}\right\}\tag{7}$$

is always valid, but the complex-valued signal $\Pi_a(t)e^{j2\pi f_0 t}$ is no analytic signal, it's just a good approximation of the analytic signal for large values of $f_0$.

Note that for $x(t)=m(t)\cos(2\pi f_0t)$ with $m(t)$ a band-limited function, i.e. $M(f)=0$ for $|f|&gt;B$, the complex-valued signal $m(t)e^{j2\pi f_0 t}$ is an analytic signal, as long as $f_0&gt;B$. The problem with the function given in your question is that $\Pi_a(t)$ is not band-limited.