but is there some other reason for calling it momentum space?
The canonical commutation relation for the position and momentum operators is (in one dimension)
$$[X, P]|\psi\rangle = (XP - PX)|\psi\rangle = i\hbar|\psi\rangle$$
On the position basis, this is
$$[x,P]\psi(x) = (xP - Px)\psi(x) = i\hbar\psi(x)$$$$[x,P_x]\psi(x) = (xP_x - P_xx)\psi(x) = i\hbar\psi(x)$$
and it follows that a position basis representation of the momentum operator is
$$P_x = -i\hbar\partial_x$$
A wavefunction with definite momentum $p$ is then
$$\psi_p(x) = e^{\frac{i}{\hbar}px} = \langle x|p\rangle$$
such that
$$P_x\psi_p(x) = -i\hbar\partial_x\,e^{\frac{i}{\hbar}px} = pe^{\frac{i}{\hbar}px} = p\psi_p(x)$$
Now, if we write
$$k = \frac{p}{\hbar}$$
then
$$\psi_p(x) = e^{ikx}$$
The momentum space wavefunction $\psi(p)$ is just the ket $|\psi\rangle$ projected onto the momentum basis:
$$\psi(p) = \langle p | \psi\rangle = \int\mathrm{d}x\, \langle p | x\rangle\langle x | \psi\rangle$$
where I have used the completeness relation
$$1 = \int\mathrm{d}x\,|x\rangle\langle x | $$ but
$$\psi(x) = \langle x | \psi\rangle$$
and
$$e^{-\frac{i}{\hbar}px} = \langle p| x\rangle$$
thus
$$\psi(p) = \int\mathrm{d}x\,\psi(x)\,e^{-\frac{i}{\hbar}px}$$
or, finally
$$\psi(k) = \int\mathrm{d}x\,\psi(x)\,e^{-ikx} $$