Uniqueness of "Limit" of Cyclic Binary Strings Set-up: By abuse, let $\sigma$ represent both the left shift operator on infinite bi-infinite strings and the cyclic left shift operator on finite strings. (Thus, for example, $\sigma(...01\bar{0}10...) = ...10\bar{1}01...$ (in the bi-infinite case, we place a bar $\bar{ }$ over the "central" character) and $\sigma(10010) = 00101$). Also let $\tilde{A}$ represent the orbit of the string $A$ under $\sigma$ in both the finite and the infinite case. For convenience let's call $A$ an "unfolding" of $\tilde{A}$ whenever $A$ is in the equivalence class $\tilde{A}$. Now let $\{\tilde{A}_i\}_{i=1}^{\infty}$ be a sequence of equivalence classes of finite binary strings.
Suppose that, for each $i$, some unfolding of $\tilde{A}_i$ is contained as a substring of some unfolding of $\tilde{A}_{i+1}$ with at least one character on either side of it; for convenience let's call this an "embedding" of $\tilde{A}_i$ into $\tilde{A}_{i+1}$. If a sequence of embeddings has been specified, then it is possible to take the "limit" of this sequence $\{\tilde{A}_i\}_{i=1}^{\infty}$ and arrive at a bi-infinite binary string $A$.
Example 1: The sequence of strings

    0
   101
  01010
 1010101
010101010
has as its limit: $...0101\bar{0}1010...$
Example 2: The sequence of strings

    1
   010
  00100
 0001000
000010000
has as its limit: $...0000\bar{1}0000...$
Conjecture: Given $\{\tilde{A}_i\}_{i=1}^{\infty}$, any sequences of embeddings lead to the same bi-infinite binary string, up to orbits of $\sigma$. That is, any two limits of this process $A$ and $B$ are related via $\sigma^k(A) = B$ for some integer $k$.
This result would be quite surprising to me if it were true, but I have not been able to construct a counterexample. It can be made more precise using the categorical language of projective limits, but I think the question is clear enough without going through this trouble.
 A: The conjecture is false.  Here is a counter-example:
Let $A_0:=\mathtt{11000}$ and $B_0:=\mathtt{11010}$ and recursively define
\begin{align}
   A_{k+1} &:= B_k B_k A_k A_k A_k \;,\\
   B_{k+1} &:= B_k B_k A_k B_k A_k \;.
\end{align}
That is, $A_n$ and $B_n$ are obtained by iterating the substitution $(\mathtt{0}\mapsto A_0, \mathtt{1}\mapsto B_0)$ on $(A_0,B_0)$.  Note that $A_k$ occurs in $A_{k+1}$ in more than one positions.  Note also that for each $n$, $A_n$ and $B_n$ do not overlap with each other or with themselves.
Let $x$ be the limit of $A_n$ as you described, by always choosing the central symbol as the "reference position":
\begin{align}
   \begin{array}[ccccc]\
      \mathtt{1} & \mathtt{1} & \underline{\mathtt{0}} & \mathtt{0} & \mathtt{0} \\
      B_0 & B_0 & \underline{A_0} & A_0 & A_0 \\
      B_1 & B_1 & \underline{A_1} & A_1 & A_1 \\
      & & \vdots & &
   \end{array}
\end{align}
In other words, $x$ has the form
\begin{align}
   \cdots\underbrace{B_2B_2\overbrace{B_1B_1\underbrace{B_0B_0\overbrace{\mathtt{11\underline{0}00}}A_0A_0}A_1A_1}A_2A_2}\cdots
\end{align}
Similarly, let $y$ be the limit of (the shifts of) $A_n$ where at each level, the fourth block is chosen as the "reference block", within which the fourth sub-block is chosen as the "reference sub-block" and so forth:
\begin{align}
   \begin{array}[ccccc]\
      \mathtt{1} & \mathtt{1} & \mathtt{0} & \underline{\mathtt{0}} & \mathtt{0} \\
      B_0 & B_0 & A_0 & \underline{A_0} & A_0 \\
      B_1 & B_1 & A_1 & \underline{A_1} & A_1 \\
      & & \vdots & &
   \end{array}
\end{align}
In other words, $y$ has the form
\begin{align}
   \cdots\underbrace{B_2B_2A_2\overbrace{B_1B_1A_1\underbrace{B_0B_0A_0\overbrace{\mathtt{110\underline{0}0}}A_0}A_1}A_2}\cdots
\end{align}
This may look different from your procedure of first doing cyclic shifts and then taking the limit of centered strings.  The limits are however the same.
Claim: $y$ cannot be a shift of $x$.
Argument: Suppose that $y=\sigma^m x$ for some $m\in\mathbb{Z}$.  Clearly $m\neq 0$. Let $n$ be such that $|A_n|=|B_n|>|m|$.  Observe that both $x$ and $y$ admit unique decompositions into blocks $A_n$ and $B_n$.  The uniqueness is because $A_n$ and $B_n$ are non-overlapping.  Since $y=\sigma^m x$, the decomposition of $y$ must be the shift of the decomposition of $x$ by $m$.  But this would require $m$ to be a multiple of $|A_n|=|B_n|$, contradicting the choice of $n$. $\Box$
Remark:
By arbitrarily choosing the third or the fourth occurrences of $A_k$ in $A_{k+1}$ as the "reference" one, we find that there are uncountably many distinct limits no two of them are shifts of each other.  I feel that this must be a special case of a standard fact about substitutions, but I am not an expert.
A: You do not need to take the cyclic shofts: the conjecture is false even without them.
Assume that $A_i$ can be embedded into $A_{i+1}$ in two ways. Then you will get $2^\omega$ possible limits; so, if they are all distinct, then you cannot hope they are in the same class. Moreover, you will get oncountably many classes.
For a concrete example, one can set $A_{i+1}=0A_i1A_i0$. If $A_i$ is embedded into $A_{i+1}$ in two different ways, its surrounding looks different, so the limits are different as well.
