I believe "maximum weight fair bipartite matching" as you've defined it is NP-hard. Even more, determining the existence of a fair bipartite matching is NP-hard.
Before I give a proof sketch, for intuition, consider the following small instance. Take $G'=(L, R, E'=L\times R)$ where $L=\{a,b\}$, $R=\{c,d,e,f\}$. Take $p$ such that $p(u,w) = 0$ for $u\in L$ and $w\in\{c,d\}$, while $p(u,w) = 1$ for $u\in L$ and $w\in\{e,f\}$. Then $a$ and $b$ are equivalent, in the sense that $p(a, w) = p(b, w)$ for all $w\in R$, so any fair matching must give $a$ and $b$ the same profit. Hence, the only fair matchings either match $a$ and $b$ to $c$ and $d$, or they match $a$ and $b$ to $e$ and $f$. Using this kind of gadget, we can force coordination of the edges in the matching. This is the basis of the reduction.
Here's an attempt at a proof. It's a bit involved. Probably there are some mistakes, but hopefully any mistakes can be fixed.
Lemma 1. Given $G'=(L, R, E'=L\times R)$ and $p:E'\rightarrow\mathbb{R}_+$ as described in the problem, determining whether $G'$ contains a fair matching is NP-hard.
Proof sketch. The proof is by reduction from Independent Set in cubic graphs. Let $(G=(V,E),k)$ be a given instance of Independent Set where $G'$ is a cubic graph (every vertex has degree 3). We describe how to construct a graph $G'=(L, R, E'=L\times R)$ and profit function $p:E'\rightarrow\mathbb{R}_+$ such that $G'$ has a fair bipartite matching if and only if $G$ has an independent set of size $k$.
The vertices in $L$ will come in pairs, called partners.
Likewise for the vertices in $R$.
For each vertex $v\in L\cup R$, we let $v'$ denote the partner of $v$.
Each vertex $\ell\in L$ and its partner $\ell'\in L$ will be equivalent, meaning that we will make
$$p(\ell, r) = p(\ell', r) \text{ for all } r\in R.$$
Consequently, any fair matching must assign the same profit to $\ell$ and $\ell'$.
In what follows, we use $\pi(\ell, r)$
to denote the value of $p(\ell,r) = p(\ell', r)$.
Further, for each pair $\ell$ in $L$,
and each pair of partners $r, r'$ in $R$,
either we make
$$\pi(\ell, r) = \pi(\ell, r')$$
or we make
$$\pi(\ell, r) \ne \pi(\ell, r').$$
In the former case, we say we allow $\ell$ and $\ell'$ to be matched to $r$ and $r'$
(because doing so would assign the same profit to $\ell$ and $\ell'$,
as required).
In the latter case, we say we prevent $\ell$ and $\ell'$ from being (both) matched to $r$ and $r'$
(because doing so would not assign the same profit to $\ell$ and $\ell'$).
As the given graph $G=(V,E)$ is cubic, it satisfies $3|V|=2|E|$, and any independent set $I$ of size $k$ in $G$ is incident to exactly $3k$ edges. Assume for ease of notation that $V=\{1,2,\ldots,n\}$.
For each edge $\{i, j\}\in E$, do the following.
Add a pair of partner vertices $r(\{i,j\}), r'(\{i,j\})$ to $R$.
For endpoint $i$, add a pair of partner vertices $\ell(i,j),
\ell'(i,j)$ to $L$. Set $$\pi(\ell(i,j), r(\{i,j\})) =
\pi(\ell(i,j), r'(\{i,j\}))= i,$$ allowing $\ell(i,j)$ and
$\ell'(i,j)$ to be matched to $r(\{i,j\})$ and $r'(\{i,j\})$.
Symmetrically, for the other endpoint $j$: add another pair of
partner vertices $\ell(j,i), \ell'(j,i)$ to $L$, and set
$$\pi(\ell(j,i), r(\{i,j\}) = \pi(\ell(j,i), r'(\{i,j\}))= j,$$
allowing $\ell(j,i)$ and $\ell'(j,i)$ to be matched to $r(\{i,j\})$
and $r'(\{i,j\})$.
For every $\ell\in L$ and $r\in R$ added so far,
if the pair $\ell, \ell'$ is not explicitly allowed (above)
to be matched to $r, r'$, then prevent the match
by assigning $\pi(\ell, r)$ and $\pi(\ell, r')$ each some unique number.
Next, add $3(|V|-k)$ pairs of filler vertices to $R$.
For each filler vertex $r$ and each $\ell(i,j)\in L$,
set $\pi(\ell(i,j), r) = 0$.
Finally, add two vertices $L_0$ and $L'_0$ (partners) to $L$, along with a two vertices $R_0$ and $R'_0$ (also partners) to $R$.
Set $\pi(L_0, R_0) = \pi(L_0, R'_0) = 1$, allowing $L_0$ and $L'_0$ to be matched to $R_0$ and $R'_0$. For every other vertex $r\in R$, set $\pi(L_0, r)$ to some unique number. (Hence, any fair matching must match $L_0$ and $L'_0$ to $R_0$ and $R'_0$.)
For every $i\in V$, for every incident edge $\{i,j\}\in E$,
set $\pi(\ell(i,j), R_0) = i$ and $\pi(\ell(i,j), R'_0) = |V|-i+1$.
That completes the reduction. To finish, we prove it is correct.
First consider for what pairs of vertices $\ell(i,j),\ell(i',j')\in L$
the latter dominates the former, that is,
$$(\forall r\in R)~\pi(\ell(i,j),r) \le \pi(\ell(i',j'), r).$$
Considering the profits assigned to edges incident to $R_0$ and $R'_0$,
this condition can only be met if $i=i'$, and, inspecting the definition of $\pi$ for the remaining edges, the condition $i=i'$ is sufficient. Hence a matching is fair if and only if it assigns $L_0$ and $L'_0$ to $R_0$ and $R'_0$, and also, for each $i\in V$, gives the same profit to all vertices in
$$N(i) = \{\ell(i,j) : \{i,j\}\in E\} \cup \{\ell'(i,j) : \{i,j\}\in E\}.$$
First, assume that $G$ has an independent set $I$ of size $k$. Obtain a fair matching for $G'$ from $I$ as follows.
Match $L_0$ and $L'_0$ to $R_0$ and $R'_0$.
For each vertex $i\in I$,
let $\{i,j_1\}, \{i,j_2\}, \{i,j_3\}$ be its three incident edges.
For each edge $\{i, j_h\}$,
match vertex $\ell(i,j_h)$ and its partner $\ell'(i,j_h)$
to $r(\{i,j_h\})$ and $r'(\{i, j_h\})$.
This gives all vertices in $N(i)$ profit $i$.
For each of the $|V|-k$ vertices $i\in V\setminus I$,
for each of the three edges $\{i,j\}$ incident to $i$,
match $\ell(i,j)$ and its partner $\ell'(i,j)$
to some unique pair of filler vertices $r$ and its partner $r'$.
This gives all vertices in $N(i)$ profit $0$.
Hence, this matching is fair.
Next, assume that $G'$ has a fair matching $M$.
$M$ must match $L_0$ and $L'_0$ to $R_0$ and $R'_0$.
For each $i\in V$, the matching must give each of the vertices in $N(i)$
the same profit. For each $\ell(i,j)\in N(i)$, its partner $\ell'(i,j)$ is also in $N(i)$. So, by inspection of the reduction,
the profit of each such vertex must be either $i$
(in which case all six vertices in $N(i)$ are matched
to vertices $r(\{i,j\})$ and their partners)
or zero
(in which case all six vertices in $N(i)$ are matched
to filler vertices in $R$).
Let $I$ be the set of vertices for which the former case holds.
For each edge $\{i,j\}$, the vertex $r(\{i,j\})$,
and its partner, are each matched to one vertex.
It follows that $I$ is an independent set.
Since the number of filler vertices is $6(|V|-k)$,
the size of $I$ must be at least $k$.
QED (?)
I think it's basically correct, if a bit convoluted. Let me know if you see any mistakes, or a way to simplify the proof.
The reduction above assumes it's okay to take $|R|>|L|$.
If that's undesirable, then I'd guess we can pad $L$
with $|R|-|L|$ filler vertices, assigning profit 0
to all of their edges except the edges to $R_0$ and $R'_0$.
We can assign profits to the latter edges to ensure
the filler vertices are not dominated by (nor dominate)
any other vertex.
