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. 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. 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).
EDIT: One has of course to specify exactly what is the language and the first-order theory, but it seems that with a sufficiently reasonable setup, the model theoretic conditions given in Scott's edit should amount to the following.
Rather than unions of sequences $\mathcal{L}$ of finite tame links $L_1\subset L_2\subset\dots$ up to ambient isotopy, we consider the sequences themselves, up to a weaker equivalence relation of pro-isotopy (defined below).
A "Fraisse link" is then the pro-isotopy class of a sequence $L_1\subset L_2\subset\dots$ satisfying the property in my comment above: whenever some $L_n$ occurs as a sublink of a finite link $L$, then this $L$ must be equivalent to a sublink of some $L_m$ relative to $L_n$ (that is, the identity on $L_n$ must extend to a self-homeomorphism of $\Bbb R^3$ sending $L$ onto a sublink of $L_m$).
Two sequences $\mathcal{L}$ and $\mathcal{L}'$ as above are pro-isotopic if each $L_i$ is equivalent to a sublink of some $L_j'$, $j=j(i)$, via a self-homeomorhpism $h_i$ of $\Bbb R^3$; and each $L_j'$ is equivalent to a sublink of some $L_k$, $k=k(j)$, via a self-homeomorphism $h_j'$ of $\Bbb R^3$, so that for each $i$, the composition $h_{j(i)}'h_i$ sends $L_i$ onto itself; and for each $j$, the composition $h_{k(j)}h_j'$ sends $L_j$ onto itself.
Pro-isotopy is indeed very similar to pro-homotopy, which brings to attention an equivalent definition of pro-isotopy (a la Pontryagin's original definition of pro-isomorphism and Siebenmann's definition of shape): pro-isotopy is the equivalence relation generated by the following two relations: 1) the relation of being a subsequence, 2) sequences $\mathcal{L}$ and $\mathcal{L}'$ are related if there exists a sequence of self-homeomorphisms $H_i$ of $\Bbb R^3$ such that $H_i(L_i)=L_i'$ and $H_{i+1}(L_i)=H_i(L_i)$.
[The proof that the two definitions of pro-isotopy are equivalent is by a standard argument: given $\mathcal{L}$ and $\mathcal{L}'$ that are pro-isotopic in the original sense, the sequence $h_1(L_1)\subset L_{j(1)}'\subset h_{k(j(1))}(L_{k(j(1))})\subset\dots$ has a subsequence that is also a subsequence of $\mathcal{L}'$; and on the other hand is related in the sense of (2) to the sequence $L_1\subset h_{j(1)}'(L_{j(1)}')\subset L_{k(j(1))}\subset\dots$, which in turn has a common subsequence with $\mathcal{L}$.]
The point of these definitions is that a Fraisse link (as defined above) is, obviously, unique.
Now that a Fraisse link is defined, I understand the original question as follows: can a Fraisse link be identified with an (ambient isotopy class of) an infinite link? The answer is no. It is clear that sequences $L_1\subset L_2\subset\dots$ and $L_1'\subset L_2'\subset\dots$ are pro-isotopic if their unions are ambient isotopic. But the converse does not hold, by the reasons that Bill Thurston gave in the end of his answer. I'm not sure that the above definitions of pro-isotopy and a Fraisse link are exactly what he had in mind, but anyhow with these definitions his argument makes sense.
