$\def\ZZ{\mathbb{Z}}\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$To spell out my comment a little more, let $Z^1$ be the sheaf of $\partial$-closed holomorphic $(1,0)$-forms. Since "holomorphic" means $\overline{\partial}$-closed, this can also be described as the space of closed $(1,0)$forms. Then we have a commutative diagram, with exact rows and columns:

$$\begin{matrix}
&& && 0 && 1 && \\
&& && \downarrow && \downarrow && \\
0 &\longrightarrow& \mathbb{Z} &\overset{2 \pi i}{\longrightarrow}& \mathbb{C} &\overset{\exp}{\longrightarrow}& \mathbb{C}^{\times} &\longrightarrow& 1 \\
& &=& &\downarrow& &\downarrow& \\
0 &\longrightarrow& \mathbb{Z} &\overset{2 \pi i}{\longrightarrow}& \cO &\overset{\exp}{\longrightarrow}& \cO^{\times} &\longrightarrow& 1 \\
& && &\phantom{\partial} \downarrow \partial& &\phantom{\partial \log}\downarrow \partial \log& \\
&& && Z^1 &=& Z^1 && \\
&& && \downarrow && \downarrow && \\
&& && 0 && 0 && \\
\end{matrix}$$
Using each of the $4$ short exact sequences, we have a diagram
$$
\begin{matrix}
H^1(\cO^{\times}) &\longrightarrow& H^2(\ZZ) \\
\downarrow && \downarrow \\
H^1(Z^1) &\longrightarrow& H^2(\CC) \\
\end{matrix}$$
and, with care, one can check that it commutes.

We also have a short exact sequence $0 \to Z^1 \to \Omega^1 \overset{\partial}{\longrightarrow} Z^2 \to 0$, where $Z^2$ is the closed $(2,0)$-forms. So we can extend this diagram to
$$
\begin{matrix}
H^1(\cO^{\times}) &\longrightarrow& H^2(\ZZ) \\
\downarrow && \downarrow \\
H^1(Z^1) &\longrightarrow& H^2(\CC) \\
\downarrow && \\
H^1(\Omega^1) && \\
\end{matrix}$$
The Atiyah class is the image of a class from $H^1(\cO^{\times})$ in $H^1(\Omega^1)$; the Chern class is the image in $H^2(\CC)$.
This much is true without assuming anything about your complex manifold.

If you want to consider the Atiyah class and the Chern class to be "the same", then it seem like you want to ask is "when does it make sense to think of $H^1(\Omega^1)$ as a subspace of $H^2(\CC)$?" If your spectral sequence degenerates at $E^1$, then the map $H^1(Z^1) \to H^2(\CC)$ is an injection and $H^1(Z^1) \to H^1(\Omega^1)$ is a surjection, so $H^1(\Omega^1)$ is a subquotient of $H^2(\CC)$. But I don't know of anything which lets you identify $H^1(\Omega^1)$ with a subspace of $H^2(\CC)$ without Hodge theory. (With Hodge theory, on a compact Kahler manifold, $H^1(Z^1)$ is the piece $H^{20} \oplus H^{11}$ in the Hodge filtration, and the maps $H^1(Z^1) \to H^2(\CC)=H^{20} \oplus H^{11} \oplus H^{02}$ and $H^1(Z^1) \to H^1(\Omega^1)=H^{11}$ are the obvious ones.)

or$0 \to \mathbb C \hookrightarrow \mathcal O \xrightarrow d \Omega^1 \to 0$ in a diagram, and assert only its commutativity, then the latter is redundant, but surely makes no claim about exactness? $\endgroup$2more comments