This is well known but not trivial. It follows primarily from Theorem 2.c.13 of [LT]. The Theorem says (applied to this situation) if every operator $T:E\to \ell_2$ is strictly singular, then every complemented subspace of $\ell_2\oplus E$ is of the form (up to isomorphism) $X'\oplus E'$ for some complemented subspace $E'$ of $E$, and $X'$ is either finite dimensional or isomorphic to $\ell_2$. So if $E'$ is finite dimensional then of course $F$ is isomorphic to $\ell_2$. In other cases, since $F$ has symmetric basis, $F$ isomorphic to $\ell_2\oplus E'$ implies, $F$ is either isomorphic to complemented subspace of $E'$ (and hence of $E$) or $\ell_2$. For the last statement see Prop 3.b.8 of [LT].

If there is an operator $T:E\to \ell_2$ which is not strictly singular, then, using the definition of not strictly singular and the fact that every subspace of $\ell_2$ is complemented, you get a complemented subspace $E'$ of $E$ isomorphic to $\ell_2$. But then $E\oplus \ell_2$ is isomorphic to $E$ by decomposition method (using $E$ has symmetric basis).

[LT] Lindenstaruss-Tzafriri, Classical Banach spaces vol 1.