Expectation of nonnegative random variable when passed through nonnegative increasing differentiable function. Part II: Electric Boogaloo
The equivalence is not true under the assumptions you have stated. The problem is that you can make $g(x)$ nearly constant on each of the intervals $[n,n+1-\epsilon)$ and then have it rapidly increase from $g(n)$ to $g(n+1)$ on the tiny interval $[n+1-\epsilon,n+1]$, which will make the derivatives $g'(n)$ contribute disproportionately to the sum.
For a concrete example of this, take $X\sim \textrm{Exp}(1)$ so that $\mathbb P(X>t)=e^{-t}$ for $t\geq 0$, and take $g(x)$ to be a smoothed and strictly increasing version of $\lfloor x\rfloor $. More precisely, let $g(x)$ be any non-negative strictly increasing differentiable function satisfying the following conditions:
- $g(x)\leq x$ for all $x\geq 0$
- $g'(n)\geq e^n$ for all $n\in\mathbb N$.
Note that these conditions do not contradict each other, since we can have $g(x)$ be nearly constant on $[n,n+1-e^{-n}]$ and then rapidly increase by nearly $1$ on an interval of size $e^{-n}$, allowing the derivative to be of size $e^n$ (or bigger). (Explicit formulas can be obtained using a quadratic spline, or smooth bump functions if so desired.)
Since $g(x)\leq x$, it follows that $\mathbb Eg(X)\leq \mathbb EX=1<\infty$. However, since $g'(n)\geq e^n$ it follows that $$ \sum_{n\in\mathbb N}g'(n)\mathbb P(X>n)\geq \sum_{n\in\mathbb N}1=\infty. $$