Product of Mittag-Leffler functions $E_\alpha(x) \cdot E_\alpha(y)$

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For the exponential function, the following relation holds: \begin{equation} e^x\cdot e^y = e^{x+y} \end{equation}

The Mittag-Leffler function \begin{equation} E_\alpha(x)=\sum_{k=0}^\infty \frac{x^k}{\Gamma(\alpha k+1)} \end{equation}

is a kind of "generalization" of the exponential function.

My question: is there a similar relation for the product of two Mittag-Leffler functions like:

\begin{equation} E_\alpha(x) \cdot E_\alpha(y) \end{equation}

or, in particular, even

\begin{equation} E_\alpha(x) \cdot E_\alpha(y) \stackrel{?}{=} E_\alpha(x+y) \end{equation}