So my understanding originally was as follows:
$$ \mathcal{F}^{-1} [\hat{f} (s)](x) = \frac{1}{2 \pi} f(-x) = \frac{1}{2 \pi} \int^\infty_{-\infty} \hat{f} (x) e^{isx} dx $$
What I'm confused on is how my inverse fourier and fourier don't match. Take, for example the function $f(x-c)$:
$$ \mathcal{F} [f(x-c)] (s) = \int^\infty_{-\infty} f(x) e^{-is(x+c)} dx = e^{isc} \hat{f} (s) $$
This is correct, and I understand how we get there. Now, if I do the inverse fourier of this, I get:
$$ \mathcal{F}^{-1} [e^{isc} \hat{f} (s)] = \frac{1}{2 \pi} \int^\infty_{-\infty} e^{isc} \hat{f} (x) e^{isx} dx = \int^\infty_{-\infty} \hat{f} (x) e^{is(x + c)} dx = \frac{1}{2 \pi} \int^\infty_{-\infty} \hat{f} (x-c) e^{isx} dx = \frac{1}{2 \pi} f(x-c) $$
Why is the $1/(2\pi)$ still there?
The $\frac{1}{2\pi}$ is there to make the Fourier transform unitary on $L^2(\mathbb{R})$. The last line of your calculation is incorrect. Your definition of the Fourier transform yields $\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(s)e^{is(x+c)}dx = f(x-c) \neq \frac{1}{2\pi}f(x-c)$. If you don't like the $\frac{1}{2\pi}$ in the definition of the inverse transform you can always use the number theorist's Fourier transform $\int_{-\infty}^\infty f(x)e^{-2\pi i sx}dx$ which has the inverse transform $\int_{-\infty}^\infty \hat{f}(s)e^{2\pi i sx}ds$.