I don't think that knot theorists are going to be very interested in such infinite links, but they do occur sometimes in the wider area of geometric topology, for instance in the proof of theorem 1.1 [here][1]. Still I doubt that they have been studied per se.

My understanding is that you are thinking about *proper* tame links with infinitely many components, i.e. such that every compact subset of $\Bbb R^3$ meets only finitely many components of the link (in general, a continuous map is called proper if the preimage of every compact set is compact). If so, then they reduce to "tame" embeddings of $X=(S^1\times\Bbb N)^+$ in $\Bbb R^3$, where $\Bbb N$ denotes the countable discrete space and $+$ stands for the one-point compactification. To see this, consider the one-point compactification of $\Bbb R^3$, which is $S^3$, and remove some distant point $pt$ (which won't make any difference since in moving the $1$-dimensional space $X$ in $S^3$ one can easily arrange to avoid $pt$).

In general, there is not so much literature about knotting of spaces more general than polyhedra; most of what is known is summarized in [the recent book by Daverman and Venema][2]. In this light the space $X$ and its "tame" embeddings don't look very attractive, to be honest.

> so realizations of this big snarly thing in $\Bbb R^3$ should all be ambient-isotopic to each other, too, but I'm not certain about that. 

Are you saying that if $L$ and $L'$ are proper tame links of countably many circles in $\Bbb R^3$ with the property that every finite link is a sublink of both $L$ and $L'$, then $L$ is ambient isotopic to $L'$?

This is not the case: take $L$ to be the union of links $L_n$, where each $L_n$ is a finite link in a ball $B_n$ of radius $1/3$ centered at $(0,0,n)$, and the ambient isotopy types of the $L_n$ are precisely those of all non-split finite links (each occurring once). On the other hand pick some infinite link $L''$ that is `non-split at infinity'; for instance such that its $i$th component has a nonzero linking number with the $(i+1)$st component for each $i$. As long as $L$ and $L''$ are disjoint, their union $L'$ is a again link that is non-split at infinity; by this I more specifically mean that it is not isotopic to any link whose $i$th component lies in the ball $B_i$ for each $i$. In particular, $L$ is not isotopic to $L'$ (even non-ambiently).

  [1]: http://front.math.ucdavis.edu/0812.1407
  [2]: http://www.calvin.edu/~venema/embeddingsbook/embeddings-noprint.pdf