That's correct: every finite quotient is solvable. Indeed let $G$ be your product of finite solvable groups. Let $p:G\to F$ be a (possibly non-continuous) surjective homomorphism to a finite group $F$. Lift $F$ to a finite subset $\tilde{F}$, and let $H$ be the closed subgroup generated by $\tilde{F}$. Then by the Nikolov-Segal theorem, the restriction of $p$ to $H$ is continuous. But as a closed subgroup of $G$, $H$ is a projective limit of finite solvable groups. So all its quotients by open subgroups are finite solvable groups.
More generally, this shows that
any abstract finite quotient of a profinite group $G$ is isomorphic to some $H/K$ where $H$ is a closed subgroup of $G$ and $K$ an open normal subgroup of $H$.
(In the particular case when $G$ is pro-$p$ for some prime $p$, this appeals to a much older theorem of Serre instead of Nikolov-Segal.)
(Edit, to answer a question in a comment.) The previous actually shows a slightly stronger statement:
(1) if $G$ is a profinite group and $F$ is a finite quotient of $G$, then there exists a t.f.g. closed subgroup $M$ of $G$ such that for every t.f.g. closed subgroup $H$ of $G$ containing $M$, $F$ is isomorphic to the quotient of $H$ by some closed normal subgroup.
[t.f.g. means topologically finitely generated, meaning having a dense finitely generated subgroup]
Now I claim that
(2) if $G$ is a product $\prod S_i$ of finite simple groups (possibly abelian), then every t.f.g. closed subgroup of $G$ is contained in a t.f.g. closed subgroup that is a product of finite simple groups.
If we combine (1) and (2), we deduce that
in an arbitrary product of finite simple groups $\prod_{i\in I}S_i$, every abstract finite quotient is isomorphic to some finite product $\prod_{j\in J}S_j$
(although it's not always given by the obvious projection!).
Actually while (1) follows from the Nikolov-Segal theorem (every finite index subgroup is open, in a t.f.g. profinite group), the particular case of the Nikolov-Segal theorem for products of finite simple group was proved by Saxl and Wilson, namely: if $(S_n)$ is a sequence of finite simple groups and $\lim_{n\to\infty}|S_n|=\infty$, then every finite index subgroup in $\prod S_n$ is open. (J. Saxl and J. S. Wilson, A note on powers in simple groups, Math. Proc. Cambridge Phil.
Soc. 122 (1997), 91-94.)
Checking (2) is quite elementary, modulo the fact that finite simple groups have a bounded generating rank, namely every finite simple group is generated by 2 elements, which relies on classification. Gather together all isomorphic simple groups in the product, to write $G=\prod_S G_S$, where each $G_S\simeq S^{I_S}$, and $S$ ranges over non-isomorphic finite simple groups. Let $L$ be a t.f.g. closed subgroup of $G$; so $L$ has a dense subgroup generated by $d$ elements for some $d$. Let $L_S$ be its projection in $S^{I_S}$; it is generated by $d$ elements. Let $M_S$ be the subgroup of $S^{I_S}$ generated by $L_S$ and the diagonal; it is generated by $\le d+2$ elements (because any finite simple group is generated by 2 elements). Also, every finitely generated subgroup of $S^{I_S}$ containing the diagonal is isomorphic to $S^k$ for some $k$. So if we choose a family of $d+2$ elements in $\prod M_S$ whose $S$-component generates $M_S$, we obtain a generating subset for a dense subgroup of $\prod M_S$, and $M_S$ is a product of finite simple groups. This proves (2).
