Re: "I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one."
Let $G$ be the the integers, and $I$ a countable indexing set. If $G^{(I)}$ were a direct summand, let $P$ be a complement summand.
We arrive at a contradiction as follows: First, $P$ is isomorphic to $G^{I}/G^{(I)}$ which contains the element $(2,4,8,16,32,ldots)$ modulo $G^{(I)}$, which (as we can peal off any of the initial terms) is a non-zero element which is infinitely divisible by 2. Second, $P$ is a subgroup of $G^{I}$, which has no infinitely divisible elements (other than zero).
I think this argument might be modified to show that the algebraic closure of a finite field will give you the counter-example you need (changing "divisibililty" to some sort of degree consideration), but I don't have a lot of time to think about it right now. I'll come back later if someone else doesn't answer your question fully.
Back now. Try the following. Let $K$ be the field obtained by adjoining to $mathbb{Q}$ the $2^{p}$th root of each prime prime $p$ (in $mathbb{Z}$). Let $G$ be the multiplicative group of $K$. Suppose by way of contradiction that $G^{(I)}$ is a direct summand of $G^{I}$, and let $C$ be a complement. As before, $C$ is isomorphic to $G^{I}/G^{(I)}$, and the element $(2,3,5,ldots )$ modulo $G^{I}$ is infinitely divisible by $2$ (thinking of ``divisibility'' multiplicatively in this case--in other words, after chopping off the front, we can take square roots as many times as we want). However, I believe it is the case that there is no element of $G^{I}$ which is infinitely divisible in this sense. (I'll leave it to the experts to prove this, but I think some form of Kummer theory would suffice. But it may be difficult to prove it.) [One last edit: I think it may even be easier to look at $mathbb{Q}(x_1,x_2,ldots)$ and adjoin a $2^{n}$th root of $x_{n}$, and modify the example accordingly.]
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