The integral of an exponential can be written as a delta function
$$\int_{-\infty}^\infty e^{ikx} dk = 2\pi\delta(x)$$
In proving this I start with the definition of a fourier transform.
$$(1)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space f(x) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty F(k) e^{ikx} dk$$
$$(2)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space F(k) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ikx} dx$$
By substituting (2) into (1),
$$(1)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space f(x) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(x')e^{-ikx'} dx'\space e^{ikx} dk$$
From there, in most proofs I've read over, they bring the exponential term $e^{-ikx'}$ out of the integrand of $dx'$ and join the two exponentials to make $e^{ik(x-x')}$.
How can the $e^{-ikx'}$ term be brought outside of the integrand of $dx'$ without changing the value of the integral?
You do not need to bring the $e^{-ikx'}$ outside of any integration; rather, you are bringing the $e^{ikx}$ inside the $dx'$ integration. Doing so, you'll get: $$ \begin{align*} f(x) &= \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(x') e^{ik(x-x')}\, \, dx' dk \end{align*} $$ Now, switch the order of integration, giving $$ \begin{align*} f(x) &= \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(x') e^{ik(x-x')}\, \, dk dx' \end{align*} $$ and then bring out $f(x')$ from the $dk$ integral: $$ \begin{align*} f(x) &= \int_{-\infty}^\infty f(x') \left(\frac{1}{2\pi}\int_{-\infty}^\infty e^{ik(x-x')}\, dk \right) dx' \end{align*} $$ and what you want is now immediate.