Is this partition problem strongly NP-complete?
Your problem can be reduced to the Partition problem (which is weakly NP-complete) without an exponential blowup of the numeric values; so your problem is weakly NP-complete, too.

This is the idea: you can view the $2N$ integers as nodes of a graph $G$, the pairs in $P$ indentify the edges between the nodes.

Clearly $G$ cannot contain cycles of odd length (checkable in polynomial time); otherwise there would be a conflict and at least one element should be excluded from the partition $A$ vs $S \setminus A$.

Each connected part $H_i$ of $G$ is bipartite $H_i = H_i' \cup H_i''$, and you can treat it like a single integer $h_i = | \sum  H_i' - \sum H_i''|$

Let $w_i$ be the difference between the number of elements in the greatest side and the number of elements in the smallest side of the bipartite component $H_i$ plus $N$: 

$w_i = N + |H_i'| - |H_i''|$ if $\sum  H_i' &gt; \sum H_i''$; 

$w_i = N + |H_i''| - |H_i'|$ otherwise.

Let $a_1,a_2,...,a_m$ be the "isolated" integers.

Let $k$ be the minimum value such that $2^k &gt; \sum h_i + \sum a_j$.

We finally build a Partition problem with the elements:

$B = \{ a_j + 2^k (N + 1) \} \cup \{ h_i + 2^k w_i \}$

It is not hard to show that $B$ has a partition $B = B_1 \cup B_2$, $\sum \{B_1\} = \sum \{B_2\}$ if and only if the original $S$ can be split into two equal size halves having the same sum and such that no pair in $P$ have both elements in the same side.

Very informally: if $h_i$ is placed on one side, it also carries the $w_i$ (positive) "weight" that keeps track of the (absolute) difference between the number of elements of $H_i'$ and $H_i''$; so on the opposite side both the sum and the weight must be "balanced" and this ensures that a solution leads to a valid equal size partition also on the original problem. In the original problem we can derive $A$ from $B_1$: if $h_i$ (or $a_j$) is placed on $B_1$ then the integer nodes of the bipartite component $H_i$ must be arranged in such  a way that $A$ recieves the elements from $H_i'$ or $H_i''$ according to which have the greater sum.

A simple example: given the integers: $1,1,2,3,4,7$ and the pairs $(7,2), (7,3)$, we build the following graph $G$:



We then reduce the connected component to a single number $(2,-1)$ (for better clarity we use the notation $(x,y)$ instead of $x + 2^k (N + y ))$; and solve the corresponding partition problem.

You can also apply the Partition problem pseudo-polynomial time dynamic programming algorithm  to the set $B$ to find the solution.