This is an extended comment to show that in the 1-dimensional case this comes down to comparing a "Hodge-theoretic inclusion" with a connecting map. 

Your version of the analytic exponential sequence implicitly rests on the map $e^{2\pi i(\cdot)}$ (giving as kernel term the constant sheaf $\mathbf{Z}$) rather than $e^{(\cdot)}$ (giving as kernel term the constant sheaf $\mathbf{Z}(1)$). The latter is more canonical insofar as it doesn't involve a choice of basis $2\pi i$ of $\mathbf{Z}(1)$, so if we work with that at the outset then in your question the division by $2\pi i$ (presumably the same choice of $\mathbf{Z}(1)$-basis as implicit in your exponential sequence) goes away, leaving us with the more "canonical" map ${\rm{d}}\log: f \mapsto {\rm{d}}f/f$.  So we'll argue in terms of this more canonical exponential sequence (and corresponding adjustment to your question).

Observe that there is an evident map from the short exact sequence
$$0 \rightarrow \mathbf{Z}(1) \rightarrow \mathscr{O}_X \stackrel{\exp}{\rightarrow} \mathscr{O}_X^{\times} \rightarrow 1$$
to
$$0 \rightarrow \mathbf{C} \rightarrow \mathscr{O}_X 
\stackrel{\rm{d}}{\rightarrow} \Omega^1 \rightarrow 1$$
(using ${\rm{d}}\log$ along the right side and the identity map on middle terms), so this yields a commutative diagram of connecting maps
$$\begin{array}[c]{ccc}
{\rm{H}}^1(X, \mathscr{O}_X^{\times}) &{\rightarrow}& {\rm{H}}^2(X, \mathbf{Z}(1)) \\
\downarrow\scriptstyle{c'_1}&&\downarrow\\
{\rm{H}}^1(X, \Omega^1_X) &{\rightarrow}& {\rm{H}}^2(X, \mathbf{C})
\end{array}$$
where the horizontal maps are connecting maps.  

Given how you defined $c_1$, your question in the case of dimension 1 thereby comes down to asking if the connecting map along the bottom coincides with the "Hodge-theoretic inclusion" (an equality in this case).  To see that such matters are not entirely trivial, note that the "Hodge-theoretic inclusion" ${\rm{H}}^0(X,\Omega^1_X) \hookrightarrow {\rm{H}}^1(X, \mathbf{C})$ is the *negative* of the connecting map arising from 
$$0 \rightarrow \mathbf{C} \rightarrow \mathscr{O}_X 
\stackrel{\rm{d}}{\rightarrow} \Omega^1 \rightarrow 1.$$
Maybe from degree 1 into degree 2 there's a cancellation of signs and one gets an equality rather than a sign discrepancy?  Good luck sorting it out.