By my comments to the question, we can assume the infinite place is not in $S$ (because its splitting in the indicated quadratic extensions is automatic). Let the elements of $S$ be $\pi_1,\dots,\pi_r$, where the $\pi_i$'s are distinct (monic) irreducibles in ${\mathbf F}_2[t]$. For each $\pi_i$, the Artin-Schreier map $\wp(x) = x^2 + x$ on the field ${\mathbf F}_2[t]/(\pi_i)$ has image equal to half of that field. Fix a nonzero element of the image, say $g_i \bmod \pi_i$ (why? you'll see shortly). For a monic irreducible $\pi$ in ${\mathbf F}_2[t]$, if $\pi \equiv 1/g_i \bmod \pi_i$ then the congruence $x^2 + x \equiv 1/\pi \bmod \pi_i$ is solvable in ${\mathbf F}_2[t]/(\pi_i)$ by the choice of $g_i$, and then by Hensel's lemma the polynomial $X^2 + X - 1/\pi$ splits in the completion ${\mathbf F}_2(t)$ at $\pi_i$, so $\pi_i$ splits in the quadratic extension of ${\mathbf F}_2(t)$ obtained by adjoining a root of $X^2 + X - 1/\pi$.

So what you would like is to find infinitely many monic $\pi$ such that $\pi \equiv 1/g_i \bmod \pi_i$ for $i = 1,\dots,r$ and $\pi$ splits completely in a predetermined finite separable extension $L/{\mathbf F}_2(t)$. We can assume $L/{\mathbf F}_2(t)$ is Galois by replacing $L$ with its Galois closure over ${\mathbf F}_2(t)$, and then the hypothesis that $\pi$ splits completely in $L$ is the same as saying $\pi$ is unramified in $L$ with trivial Frobenius conjugacy class in the Galois group of $L/{\mathbf F}_2(t)$.

At the same time, the congruence condition $\pi \equiv 1/g_i \bmod \pi_i$ for monic irred. $\pi$ is also a Frobenius constraint using Carlitz extensions. The group $({\mathbf F}_2[t]/(\pi_i))^\times$ is naturally the Galois group of the splitting field over ${\mathbf F}_2(t)$ of the Carlitz polynomial associated to $\pi_i$. Write this splitting field as $K_{\pi_i}$. The natural isomorphism of the (abelian) Galois group of $K_{\pi_i}/{\mathbf F}_2(t)$ with $({\mathbf F}_2[t]/(\pi_i))^\times$ identifies the Frobenius element attached to $\pi$ with $\pi \bmod \pi_i$ for any monic irred. $\pi$ in ${\mathbf F}_2[t]$ not equal to $\pi_i$. (This is analogous to the way $({\mathbf Z}/(m))^\times$ is identified with the Galois group of a cyclotomic field with the Frobenius at a prime $p$ being $p \bmod m$.) Therefore asking that the monic irred. $\pi$ satisfy $\pi \equiv 1/g_i \bmod \pi_i$ for all $i$ is a bunch of simultaneous Frobenius conditions on $\pi$ in the fields $K_{\pi_1},\dots,K_{\pi_r}$. (Warning: these Frobenius conditions for $\pi$ are usually *not* conditions for $\pi$ to split completely in $K_{\pi_i}$ because we're working with *reciprocals* of nonzero elements mod $\pi_i$ that are in the image of the Artin-Schreier map, and the Artin-Schreier map has no good multiplicative properties. Maybe you could get a split completely interpretation using the reciprocal of a nonzero element in the image of the Artin-Schreier map mod $\pi_i$ which is itself in the image of the Artin-Schreier map mod $\pi_i$. A count shows it is very likely that there is such an overlap, but I haven't checked it for certain.)

The Carlitz extensions $K_{\pi_1},\dots,K_{\pi_r}$ are linearly disjoint over $F_2(t)$, so there is no problem finding infinitely many $\pi$ with given Frobenius elements in each of the Galois groups of $K_{\pi_i}/{\mathbf F}_2(t)$. (This is also a consequence of the Chinese remainder theorem and Dirichlet's theorem modulo $\pi_1\cdots\pi_r$.) *If* the field $L$ is linearly disjoint from all the fields $K_{\pi_i}$ over ${\mathbf F}_2(t)$ then we can express all of Pete's original conditions as a single Frobenius condition in the Galois group of their composite field over ${\mathbf F}_2(t)$ and that has infinitely many solutions by Chebotarev. On the other hand, if $L$ is not linearly disjoint from those Carlitz extensions over ${\mathbf F}_2(t)$ then there could be some subtle compatibility issues to be sure Chebotarev can be applied. I'll stop at this point since I've shown the basic plan for how to interpret everything as a Frobenius condition in a Galois group over ${\mathbf F}_2(t)$.

monicnonconstant $h$ in ${\mathbf F}[t]$. (The polynomial $X^2+X-1/h$ is irreducible over ${\mathbf F}(t)$ since it's quadratic and one can check by contradiction that this polynomial has no root in ${\mathbf F}(t)$ because $h$ is nonconstant in ${\mathbf F}[t]$.) – KConrad May 18 '11 at 5:27