Consider the sentence $P:\forall x\exists y:y^3=1\neq y,yx=xy$. Let $(F_i)$ be an infinite family of groups, each of which possesses an element of order 3. Then $\bigoplus F_i$ satisfies $P$.

Now assume that each $F_i$ is non-abelian of order 6: in $F_i$, elements of order 2 never commute to elements of order 3. Thus in the direct product, an element all of whose components have order 3 commutes with no element of order 2. This answers (1).

Also when each $F_i$ is non-abelian of order 8, the direct sum satisfies ``each element has a non-abelian centralizer" (which can be made 1st order), but not the direct product.

More generally, if there exists $n_0$ such that in each $F_i$, there exists a family of $n_0$ elements whose centralizer in $F_i$ is abelian (this obviously holds if $F_i$ has bounded cardinal, but also more generally if $F_i$ has finite generating rank (e.g. if $F_i$ is simple since then it is generated by 2 elements). Then the direct product $\prod F_i$ contains a finite family whose centralizer is abelian; while the direct sum $\bigoplus F_i$ has such a finite family only if all but finitely many $F_i$'s are abelian. In this very general case again, the direct sum and the direct product are not elementary equivalent, and when $F$ is finite non-abelian no infinite $F^{(\alpha)}$ can be elementary equivalent to any $F^\kappa$.

Since I don't know a single case for which infinitely of the $F_i$'s are non-abelian and I know the direct sum to be elementary equivalent to the direct product, I have no idea about (2).

*Edit:* It took me some time to struggle until I solved the following: in a group $F$, let $m(F)$ be the smallest cardinal of a subset of $F$ whose centralizer is abelian. **Find finite groups with $m(F)$ arbitrary large** (even finding $F$ with $m(F)\ge 3$ did not seem immediate). Note that $m(A\times B)=\max(m(A),m(B))$, so direct products do not help; $m(F)$ is bounded above by the minimal number of generators, and is 0 for abelian groups. Actually, if we fix a prime $p$ and consider the product of a large family of $n$ non-abelian groups of order $p^3$ and glue their center, so that the resulting group $G$ has order $p^{2n+1}$ and has both its center and its derived subgroup of order $p$, then the centralizer of any noncentral element has index $p$, and hence the centralizer of any family of $k$ elements has index $\le p^k$; on the other hand this group has no abelian subgroup of order $>p^{n+1}$, so $m(G)\ge n+1$ (actually it's an equality). This provides families of $(F_i)$ for which the previous argument does not work.