I believe the answer to the question in the title is probably **Cauchy**, who in *Méthode simple et générale pour la détermination numérique des coefficients que renferme le développement de la fonction perturbatrice*, C. R. Acad. Sci. Paris **11** (1840), [453-475][1], writes (page [469][2]):


"La formule
$$
Q = \sum Q_{h,h'} e^{h(p-\varpi)\sqrt{-1}} e^{h'(p'-\varpi')\sqrt{-1}}
$$
entraîne l'équation
$$
Q_{h,h'} = \frac1{4\pi^2} \int_0^{2\pi} \int_0^{2\pi} Q e^{-h(p-\varpi)\sqrt{-1}} e^{-h'(p'-\varpi')\sqrt{-1}} dp\\,dp'."
$$
Specializing to $\varpi=\varpi'=0$ one gets the requested complex Fourier series and formula for its coefficients.

[**Edit:** H. Burkhardt in *Trigonometrische Reihen und Integrale*, Encykl. Math. Wiss. II A 12, p. [929][3] confirms the above as the first of several papers where Cauchy uses the complex form. Moreover he goes back even further to **Laplace** who writes in *Théorie analytique des probabilités* (Paris, 1812), pp. [83-84][4]:

"Take the equation $u=\sum_{x=0}^\infty y_xt^x$. Substitute on both sides $e^{x\varpi\sqrt{-1}}$ for $t^x$... and write $U$ for what $u$ then becomes. Multiplying the equation by $e^{-x\varpi\sqrt{-1}}$ and integrating... the right-hand side boils down to $2\pi y_x$; one has therefore $y_x = \frac1{2\pi}\int U\ d\varpi\ (\cos x\varpi  - \sqrt{-1}\sin x\varpi)$".]


Note also that *specific* complex Fourier series were written much earlier. For instance Lagrange in [1766][5] computes
$$
(1-\alpha e^{i\theta})^{-s} = \sum_{m=0}^{\infty}\binom{-s}{m}(-\alpha)^me^{im\theta}
$$
... except  that he still writes everywhere $e^{im\theta}$ in the form $\cos m\theta +\sin m\theta\sqrt{-1}$.


  [1]: http://gallica.bnf.fr/ark%3A/12148/bpt6k2970g.f453
  [2]: http://gallica.bnf.fr/ark%3A/12148/bpt6k2970g.f469
  [3]: http://archive.org/stream/p1n2encyklopdied02akaduoft#page/929/
  [4]: http://archive.org/stream/oeuvrescomplte07lapluoft#page/84
  [5]: http://archive.org/stream/uvresdelagrange01lagr#page/620/