It is possible that $\dim_F(E_1)=\dim_F(E_2)=0$ yet $\dim_F(E_1+E_2)=1$, so there is no inequality in the opposite direction.
In fact, Falconer's example of sets $E_1, E_2$ such that $\dim_H(E_1)=\dim_H(E_2)=0$ but $\dim_H(E_1+E_2)=1$ already works. In Falconer's example, not only $E_1+E_2$ has dimension $1$, but in fact $E_1+E_2$ is an interval. Hence $\dim_F(E_1)=\dim_F(E_2)=0$ (since $\dim_F$ is bounded above by $\dim_H$) but $\dim_F(E_1+E_2)=1$.
There are also examples which are not forced by the Hausdorff dimensions of the sets. Indeed, let $C$ be the ternary Cantor set. It is a classical result of Kahane and Salem that if $\mu$ is any measure supported on $C$, then $\widehat{\mu}(\xi)\nrightarrow 0$ as $|\xi|\to\infty$; in particular, $\dim_F(C)=0$. Clearly, the same is true for any dilate $t C$ with $t\neq 0$. Now since $C\times C$ has dimension $2\log 2/\log 3>1$, Marstrand's theorem on projections tells us that $C+ tC$ has positive Lebesgue measure for almost every $t$. Hence, for almost every $t$ we have $\dim_F(C)=\dim_F(tC)=0$ but $\dim_F(C+ tC)=1$.
Morally speaking, there is no reason why $\dim_F(E_1+E_2)=\min(1,\dim_F(E_1)+\dim_F(E_2))$ should hold. Leaving Hausdorff dimension considerations aside, if $E_1$ or $E_2$ are not Salem, this tells us that there are some resonances in the construction of the sets at a set of frequencies (possibly very sparse). These special frequencies will in general be lost in the sum $E_1+E_2$ (unless $E_1$ and $E_2$ also resonate to each other in some strong form), so one would expect that $\dim_F(E_1+E_2) > \dim_F(E_1)+\dim_F(E_2)$. However, I suspect it is not trivial at all to give specific examples where $\dim_H(E_1+E_2)<1$ (because proving Salemness or even some good decay of Fourier coefficients is usually hard).