Let $S$ be a family of functions $\mathbb{R} \to \mathbb{R}$ that are uniformly bounded. Assume that the functions of $S$ are in $L^1(\mu)$, where $\mu = e^{-2|x|}L$, where $L$ is the Lebesgue measure. How can I prove that $u:\mathbb{R} \to \mathbb{R}$ such that $$u(x) = \limsup_{y \to x} \sup\{v(y): v \in S\}$$ is also in $L^1(\mu)$ if we assume that $\sup\{v(y): v \in S\} < \infty$ and $u(x) = \limsup\limits_{y \to x} \sup\{v(y): v \in S\} < \infty$?
I think that the statement is true. Or, at least, I've not been able to produce a counterexample (under these assumptions). The problem arised while working on a generalization of the dominated convergence theorem (but the origin is not relevant to the statement).
What you want to show is false.
As a counterexample, consider the set $S = \{f_n : n \in \Bbb{N}\}$ with $f_n(x) = 1_{(1,n)}(x)/x$.
I leave it to you to verify $u(x) = 1_{[1,\infty)}(x) /x$, which is not in $L^1$.
With the updated form of the question, take $f_n (x) = e^{2x} 1_{(1,n)}(x)/x$, so that $u(x) = e^{2x} 1_{[1,\infty)}(x)/x$.
With the updated updated answer, the answer is true: It is not hard to see that $\mu$ is a finite measure. Furthermore, the family $S$ is now uniformly bounded, say $|v(x)|\leq C$ for all $v \in S$ and $x\in \Bbb{R}$. This easily(!) implies $|u(y)|\leq C$. Since $\mu$ is a finite measure, this implies $u \in L^1$.
Finally: the above is only true if $u$ is measurable. For general (uncountable) $S$, this might fail!!