The correct way to pin this down is to look at the full, time-dependent exponential. You can choose to describe a wave moving towards $+x$ either as
\begin{align}
\Psi(x,t)&=e^{i(kx-\omega t)}\, ,\tag{1} \\
\hbox{ or }\quad \Psi(x,t)&= e^{i(\omega t-kx)}\, .\tag{2}
\end{align}
Whichever you choose then determines the sign of the Fourier transform and its inverse. Assuming (1) for the purpose of the example, then the spatial part of a plane wave moving in the $+x$ direction would be
$$
\psi(x)=\langle x\vert p\rangle=N e^{ipx/\hbar}\, . \tag{3}
$$
with $N$ a normalization factor. The rest is basically using the completeness relation. If, for symmetry reason, one sets
$$
\hat I=\int dx\,\vert x\rangle \langle x\vert =
\int dp\,\vert p\rangle \langle p\vert \tag{4}
$$
(which I think is easier to remember) then
\begin{align}
\langle p\vert \psi\rangle = \psi(p)&=
\int dx\,\langle p\vert x\rangle \langle x\vert\psi\rangle\, ,\\
&= \int dx N^* e^{-ipx/\hbar}\psi(x)\, , \tag{5}
\end{align}
Note that in (5) I've explicitly used $\langle p\vert x\rangle = N^*e^{-ipx/\hbar}$, which will pin down the $N$ through
\begin{align}
\langle x\vert \psi\rangle = \psi(x)&=
\int dp\,\langle x\vert p\rangle \langle p\vert\psi\rangle\, ,\\
&= \int dp N e^{ipx/\hbar}\psi(p)\, ,\\
&= \int dp dx N N^* \psi(x)\, ,\\
\end{align}
With this you find $NN^*=1/(2\pi\hbar)$. The choice of $N=1/\sqrt{2\pi \hbar}$ makes the transformation between the direct and inverse transformation very symmetric since
$$
\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}
=\langle p\vert x\rangle^*\, .
$$
