If you have a complex baseband signal
$$x(t)=x_R(t)+jx_I(t)=|x(t)|e^{j\phi(t)}\tag{1}$$
and you multiply it with a (real-valued) sinusoid, the resulting signal is obviously still complex. What happens in IQ-modulation is that you generate a real-valued band-pass signal containing the same information as $x(t)$:
$$s(t)=\text{Re}\{x(t)e^{j\omega_ct}\}=x_R(t)\cos(\omega_ct)-x_I(t)\sin(\omega_ct)\tag{2}$$
Using the magnitude and phase of $x(t)$, the bandpass signal $s(t)$ can also be written in the form
$$s(t)=|x(t)|\cos[\omega_ct+\phi(t)]\tag{3}$$
which shows that in general $s(t)$ is amplitude and phase modulated.
EDIT:
This is what happens in the frequency domain when you use a sinusoid for modulation as opposed to IQ modulation. Let
$$r(t)=x(t)\cos(\omega_ct)=x(t)\frac12\big[e^{j\omega_ct}+e^{-j\omega_ct}\big]\tag{4}$$
In the frequency domain this corresponds to
$$R(j\omega)=\frac12\big[X(j(\omega-\omega_c)+X(j(\omega+\omega_c))\big]\tag{5}$$
For the IQ modulated signal $s(t)$ we get (from $(2)$)
$$s(t)=\text{Re}\{x(t)e^{j\omega_ct}\}=\frac12\big[x(t)e^{j\omega_ct}+x^*(t)e^{-j\omega_ct}\big]\tag{6}$$
The Fourier transform of $(6)$ is
$$S(j\omega)=\frac12\big[X(j(\omega-\omega_c)+X^*(-j(\omega+\omega_c))\big]\tag{7}$$
Comparing $(5)$ and $(7)$ you see that the part of the spectrum that is centered around $-\omega_c$ is inverted for $s(t)$ (i.e., it is a mirror image of $X(j\omega)$; note the negative sign before $j$), whereas it is just shifted without inversion for the cosine modulated signal $r(t)$.
