Yet another definition of Lebesgue integral

Yes. Note that $s(f,P)$ and $S(f,P)$ are the integrals of simple functions. These simple functions converge pointwise to $f$ along sequences archiving the supremum and infimum respectively. Furthermore, they are dominated by $M$, so the result follows from the dominated convergence theorem.

Details:

We have $s(f,P)=\sum_{i= 1}^k=\mu(A_i)t_{i-1}=\int\sum_{i= 1}^k 1_{A_i}t_{i-1}$, the integral of a simple function. Moreover, every such simple function is bounded in $[0,M]$ and is therefore dominated by the constant function with value $M$, which is integrable since we re working with a finite measure space. By the monotone convergence theorem, it suffices to show that there is a sequence $(P_n)$ such that $\lim_{n\to\infty} s(f,P_n)=s$ and the corresponding simple functions converge pointwise to $f$ (the case with the other approximation works similarly).

Note that if $P$ refines $P'$, then $s(f,P)\geq s(f,P')$. So we can take any sequence $(P_n)$ such that $\lim_{n\to\infty} s(f,P_n)=s$ and by refining each partition make sure that it converges in the same manner but such that no interval in the underlying partition is longer than $1/n$. But this means that the corresponding simple function is never more than $1/n$ different from $f$, so we can make the underlying simple functions converge to $f$.