I am familiar with both of the following Fourier transformation identities:
$$e^{jw_0t} \leftrightarrow 2\piδ(w-w_0)$$ and
$$x(t)e^{jw_0t} \leftrightarrow X(w-w_0)$$
However I need to find out why the $2\piδ$ disappears from the frequency domain when the input signal is multiplied by $x(t)$.
Please could anyone elucidate as to why this is?
My existing knowledge on signal processing is inferior to many people on here, please bare that in mind. Any help is greatly appreciated.
The Convolution Theorem of the Fourier Transform and the properties of the Dirac Delta under integration can hopefully shed some light for you:
$$\begin{align}F\left\{x(t)e^{j\omega_0t}\right\} &= \dfrac{1}{2\pi}\left(F\left\{x(t)\right\}*F\left\{e^{j\omega_0t}\right\}\right) \\ \\ &= \dfrac{1}{2\pi}\left[X(\omega) *2\pi\delta(\omega-\omega_0)\right]\\ \\ &= \int_{-\infty}^{\infty} X(\omega-\tau)\delta(\tau-\omega_0)\space d\tau\\ \\ &= X(\omega-\omega_0)\\ \end{align}$$
Multiplication of two functions in the time domain is a convolution of their spectra in the frequency domain.
Multiplication doesn't make the Dirac Delta function "disappear"; multiplication performs a convolution in the frequency domain.
The Dirac Delta contributes to the final result of the convolution. Its contribution specifically results in the shift of $x(t)$'s frequency spectrum, $X(\omega)$, to a new origin, $\omega_0$.
Update
Convolution is defined as
$$f(\omega) * g(\omega) = \int_{-\infty}^{\infty} f(\omega-\tau)g(\tau) \space d\tau$$
To proceed with this specific situation, just use:
$$f(\omega) = X(\omega)$$
$$g(\omega) = 2\pi\delta(\omega -\omega_0)$$