Consider a sequence of random variables $(X_i)_{i\geq 1}$, with probability messures $(P_i)_{i\geq 1}$, and a bounded deterministic sequence $(a_i)_{i\geq 1}$ satisfying
$$ \lim_{n \to \infty}P_n(X_i/a_i>1)=1. $$
Let $(E_i)_{i \geq1}$ be a generic sequence of $P_i$-measurable events: can we claim that there exists $n_0 \in \mathbb{N}$ such that, for all $n \geq n_0$,
$$ a_n P_n(E_n)=\int_{E_n}a_nP_n(dx_n)\leq \int_{E_n}x_n P_n(dx_n)=\mathbb{E}_{P_n}(X_n\boldsymbol{1}_{E_n}), $$ where $\boldsymbol{1}_{E_n}$ denotes the indicator function of the set $E_n$ and $\mathbb{E}_{P_n}$ the expectation with respect to $P_n$? Which additional conditions would be eventually needed to claim it? Would the answer change if the event $\liminf_{n \to \infty}\{X_i/a_i>1\}$ received probability 1 (i.e. if the inequality holds true ultimately almost surely)?
No, this doesn't hold. Here's a basic example: let $a_n = 1$ for all $n$. Let $X_n = 2 + \xi_n$ where $\xi_n$ is $-2n$ with probability $1/n$ and $0$ otherwise. Then the hypothesis is satisfied, but if we take $E_n = \Omega$, the inequality fails, since $\mathbb{E}[X_n] = 0$.
Some sort of uniform integrability statement is necessary.
EDIT: suppose we add that the collection $\{X_n / a_n\}$ is uniformly integrable. Then we could write $$ \mathbb{E}[(X_n/a_n) \mathbf{1}_{E_n}] = \mathbb{E}[(X_n/a_n) \mathbf{1}_{E_n} \mathbf{1}_{X_n > a_n}] + \mathbb{E}[(X_n/a_n) \mathbf{1}_{E_n} \mathbf{1}_{X_n \leq a_n}]\,. $$
The latter expectation converges to zero as $n \to \infty$ since the variable $(X_n/a_n) \mathbf{1}_{X_n \leq a_n}$ converge to $0$ almost surely and are uniformly integrable. So we get $$ \mathbb{E}[(X_n/a_n) \mathbf{1}_{E_n}] = \mathbb{E}[(X_n/a_n) \mathbf{1}_{E_n} \mathbf{1}_{X_n > a_n}] + o(1) \geq \mathbb{E}[\mathbf{1}_{E_n} \mathbf{1}_{X_n > a_n}] + o(1) = \mathbb{P}(E_n) + o(1)\,. $$
I think it is unlikely that you will be able to get a genuine inequality (i.e. with the $o(1)$ term replaced by $0$ just using assumptions like uniform integrability, unless more assumptions on $E_n$ are made, or there is a bit of extra room in the inequalities $X_n > a_n$.