Solution:
First, assume the 2-configuration has just one cell. We claim there exists a sequence $a_{0}, a_{1}, \ldots, a_{n}$ of elements of $A$ (not necessarily all distinct) such that the list
$$\{a_{0}, a_{1}\}, \{a_{1}, a_{2}\}, \ldots, \{a_{n-1}, a_{n}\}, \{a_{n}, a_{0}\}$$
contains each element of the 2-configuration exactly once. To see this, consider the longest sequence such that $\{a_{0}, a_{1}\}, \ldots, \{a_{n-1}, a_{n}\}, \{a_{n}, a_{0}\}$ are all distinct elements of the 2-configuration. (We may take $n=0$ if necessary. Note that the finiteness condition ensures such a maximal sequence exists.) Each element of $A$ occurs an even number of times among these pairs (since each occurrence in the sequence contributes to two pairs). If every element occurs 4 times or 0 times, then the elements occurring in the sequence form a cell, since they cannot occur in any other pairs in the 2-configuration. Hence, they are all of $A$, and our sequence uses all the pairs in the 2-configuration, so the claim follows. Otherwise, there is some element $a_{i}$ occurring exactly twice. Choose $b_{1}$ so that $\{a_{i}, b_{1}\}$ is one of the two pairs in the 2-configuration not used by our sequence. Then choose $b_{2}$ so that $\{b_{1}, b_{2}\}$ is another pair not used thus far. Continue in this manner, choosing new elements $b_{k}$ with $\{b_{k}, b_{k+1}\}$ a pair not already used, until we reach a point where finding another unused pair is impossible. Now, our pairs so far are
$$\begin{gathered} \{a_{0}, a_{1}\}, \ldots, \{a_{n-1}, a_{n}\}, \{a_{n}, a_{0}\}, \\ \{a_{i}, b_{1}\}, \{b_{1}, b_{2}\}, \ldots, \{b_{k-1}, b_{k}\} . \end{gathered}$$
Every element is used in an even number of these pairs, except possibly $a_{i}$, which is used in three pairs, and $b_{k}$, which is used in an odd number of pairs (so one or three) unless $a_{i}=b_{k}$, in which case this element occurs four times. But since it is impossible to continue the sequence, $b_{k}$ must indeed have been used four times, so $b_{k}=a_{i}$.
But now we can construct the following sequence of distinct elements of the 2-configuration:
$$\begin{gathered} \{a_{0}, a_{1}\}, \ldots, \{a_{i-1}, a_{i}\}, \{a_{i}, b_{1}\}, \{b_{1}, b_{2}\}, \ldots, \{b_{k-1}, a_{i}\}, \\ \{a_{i}, a_{i+1}\}, \ldots, \{a_{n-1}, a_{n}\}, \{a_{n}, a_{0}\} . \end{gathered}$$
This contradicts the maximality of our original sequence. This contradiction means that our original sequence must have used all the pairs in the 2-configuration, after all.
So we can express the 2-configuration via such a sequence of pairs, where each pair's second element equals the first element of the next pair. If $A$ has $n$ elements, then (since each element appears in four pairs) we have $2n$ pairs. So we can choose the 1st, 3rd, 5th, $\ldots, (2n-1)$th pairs, and then each element of $A$ belongs to just two of these pairs, because each occurrence of the element as an $a_{i}$ contributes to two consecutive pairs from our original sequence (or the first and last such pairs). Thus, we have our consistent 2-configuration of order 2, as desired.
Finally, if $A$ consists of more than one cell, then the pairs within any given cell form a consistent 2-configuration of order 4 on that cell. So we simply apply the above procedure to obtain a consistent 2-configuration of order 2 on each cell, and then combining these gives a consistent 2-configuration of order 2 on $A$, as desired.