Phasors and Fourier Transform
Firstly, your  phasor definition should have an \$ e^{jwt}\$ that is usually omitted as we generally deal with a single frequency system and can hence divide out that term.
The only remaining difference between the two is then that the Fourier transform one has a delta function at the positive AND negative frequencies which is missing from the phasor.
This is because a phasor is the simplest example of an analytic signal.
The idea with an analytic signal is that for a real valued function its Fourier transform's amplitude spectra will always be even, i.e. symmetric about the origin. Similarly, the phase spectra is odd, i.e. anti-symmetric about the origin. Therefore, it is pointless to involve both the positive and negative frequencies in calculations because one tells us all the information about the other.
An analytic signal is defined as:
\$ s_a(t) = s(t) + js_h(t)\$
Where \$ s_h(t) \$ is the hilbert transform of the original signal \$ s(t) \$
For a cosine \$ cos(\omega t + \phi)\$
\$ 
s_a(t) = cos(\omega t + \phi) + jsin(\omega t + \phi) \\
s_a(t) = e^{j(\omega t + \phi)} \\
s_a(t) = e^{j\omega t}e^{j\phi}
\$
Which is our familiar phasor, without the time term divided out.
It turns out the spectrum of this analytic signal is a scaled version of the original signal's spectrum but WITHOUT any negative frequencies (There's a factor of 2 carried around the place that I will omit for simplicity, see the link for the details).
So the Fourier transform of our analytic signal will have only one positive Delta function, which simplifies calculations.
It is not strictly correct then that the phasor is equal to the Fourier transform but rather they are closely related. It also turns out that to bring back the negative frequencies all we have to do is take the real part of the analytic signal, which is what we do when converting a phasor back to a cosine!
Analytic signals are even more useful in converting a bandpass signal or system to an equivalent baseband form.