Identifying the product of two Fourier series with a third?

I'll assume from the text that you're using the asterisk to denote multiplication. This is a bit confusing since in the context of convolutions it is usually used to denote convolution, so I'll use an asterisk to denote convolution and a dot to denote multiplication.

You're right, this can be expressed as a convolution, as follows:

$$ \begin{eqnarray} \sum_k{c_k \mathrm e^{\mathrm i k t}} \cdot \sum_{k'}{c'_{k'} \mathrm e^{\mathrm i {k'} t}} &=& \sum_{k,k'}{c_k c'_{k'}\mathrm e^{\mathrm i k t}} \mathrm e^{\mathrm i {k'} t} \\ &=& \sum_{k,k'}{c_k c'_{k'}\mathrm e^{\mathrm i (k+k') t}} \\ &=& \sum_{k,k''}{c_k c'_{k''-k}\mathrm e^{\mathrm i k'' t}} \\ &=& \sum_{k''}\left(\sum_kc_k c'_{k''-k}\right)\mathrm e^{\mathrm i k'' t} \\ &=& \sum_{k''}\left(c*c'\right)_{k''}\mathrm e^{\mathrm i k'' t} \\ &=& \sum_{k''}c''_{k''}\mathrm e^{\mathrm i k'' t} \end{eqnarray} $$

with $c''_{k''}=\left(c*c'\right)_{k''}$.

I'd use the notation $ \times $ rather than $ * $ because the latter is used for convolutions in this sort of context (Fourier analysis). In any case, you can explicitly calculate the coefficients of the product's Fourier series via $$ c''_n = \sum_{k=-\infty}^{\infty} c_{n-k} c'_k$$ Note that this can be related to convolutions in the sense that $ c''_n = (c * c')_n $.