$\def\cO{\mathcal{O}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}$Yes, this is right. This answer will prove the equality of three maps $H^1(\cO^{\ast}) \to H^2(\CC)$.

(a) The map $$H^1(\cO^{\ast}) \overset{c_1}{\longrightarrow} H^2(\ZZ) \overset{2 \pi i}{\longrightarrow} H^2(\CC)$$ induced from the exponential sequence and the inclusion of $\ZZ$ into $\CC$.

(b) The map $$H^1(\cO^{\ast}) \overset{d \log}{\longrightarrow} H^1(W^1) \to H^1(Z^1) \cong H^2(\CC)$$ where $W^1$ is the $d$-closed $(1,0)$-forms and $Z^1$ is the $d$-closed $1$-forms.

(c) The map $$H^1(\cO^{\ast}) \overset{d \log}{\longrightarrow} H^1(\Omega^1) \cong H^{11} \to H^2_{DR} \cong H^2(\CC)$$ where $H^{11}$ is the harmonic $(1,1)$-forms.

We will use the following notations:

$A^{pq}$ the sheaf of smooth $(p,q)$-forms.

$W^p$ the sheaf of $d$-closed $(p,0)$ forms.

$\Omega^p$ the sheaf of holomorphic $(p,0)$ forms.

$A^k$ the sheaf of $k$-forms.

$Z^k$ the sheaf of closed $k$-forms.

$H^{pq}$ are the harmonic $(p,q)$ forms.

We have a commutative diagram with exact rows:
$$\begin{matrix}
0 & \to & \ZZ & \overset{2 \pi i}{\longrightarrow} & \cO & \overset{\exp}{\longrightarrow} & \cO^{\ast} & \to & 0 \\
& & 2 \pi i \downarrow & & = & & \downarrow d \log & & \\
0 & \to & \CC & \longrightarrow & \cO & \overset{d}{\longrightarrow} & W^1 & \to & 0 \\
& & = & & \downarrow & & \downarrow & & \\
0 & \to & \CC & \longrightarrow & C^{\infty} & \overset{d}{\longrightarrow} & Z^1 & \to & 0 \\
\end{matrix}$$
Exactness of the first row is that nonvanishing holomorphic functions locally have holomorphic logarithms, the last row is the Poincare lemma, and the second row is the Poincare lemma plus the fact that integrals of holomorphic functions are holomorphic.

The boundary map on sheaf cohomology is functorial in such diagrams of short exact sequences, so we have a commutative diagram:
$$\begin{matrix}
H^1(\cO^{\ast}) & \overset{c_1}{\longrightarrow} & H^2(\ZZ) \\
d \log \downarrow & & \downarrow 2 \pi i \\
H^1(W^1) & \longrightarrow & H^2(\CC) \\
\downarrow & & = \\
H^1(Z^1) & \cong & H^2(\CC) \\
\end{matrix}$$
The last row is an isomorphism by partitions of unity.

The equality (a)=(b) expresses two ways of going around this diagram.

To relate (b) and (c), we need some generalities about $H^q(W^p)$ which I wish I knew a citation for.

We first recall that the Poincare short exact sequence
$$0 \to Z^p \to A^p \overset{d}{\longrightarrow} Z^{p+1} \to 0$$
induces isomorphisms
$$H^{k}_{DR} = H^0(Z^k)/d H^0(A^{k-1}) \cong H^1(Z^{k-1}) \cong \cdots \cong H^k(Z^0) = H^k(\CC) .$$

**Key compatability:** The map $W^p \to Z^p$ induces an inclusion $H^q(W^p) \to H^q(Z^p) \cong H^{p+q}(\CC)$ with image $\bigoplus_{k \leq q} H^{(p+q-k)k}$. The map $H^q(W^p) \to H^q(\Omega^p)$ induces the projection $\bigoplus_{k \leq q} H^{(p+q-k)k} \to H^{pq} \cong H^q(\Omega^p)$.

Using this, we prove (b)=(c). The image of $H^1(W^1)$ in $H^1(Z^1)$ is $H^{20} \oplus H^{11}$, so every class in $H^1(\cO^{\ast})$ is sent by (b) to a class of the form $\alpha \oplus \beta$. As you say, we can use the description from (a) to show that this class is realizable by a $(1,1)$-form, so $\alpha=0$. Then $\beta \in H^{11}$ is the image of the composition $H^1(W^1) \to H^2(\CC) \to H^{11}$, and the key compatability says that this is the same as the map $H^1(W^1) \to H^1(\Omega^1)$. So we can compute $\beta$ using the map $H^1(W^1) \to H^1(\Omega^1)$, and that is your option (c).

I'm going to hold off on writing a proof of the Key Compatability on the assumption that someone will give me a citation for it, ideally in the same sort of classical language that the OP uses in the question.