With respect to $J$ we get a splitting of the complexified differential forms; namely,
$$\Omega^k(X)\otimes\mathbb{C} = \bigoplus_{p+q=k}\Omega^{p,q}(X).$$
With respect to this splitting, we have the operator $\overline{\partial} : \Omega^{p,q}(X) \to \Omega^{p,q+1}(X)$ given by $\pi^{p,q+1}\circ d$ where $d$ is the exterior derivative and $\pi^{p,q+1}$ is the projection $\Omega^{k+1}(X)\otimes\mathbb{C} \to \Omega^{p,q+1}(X)$.
Integrability of $J$ is equivalent to the condition $\overline{\partial}^2 = 0$, but this is automatic on a real surface because we have
\begin{align*}
\Omega^0(X)\otimes\mathbb{C} &= \Omega^{0,0}(X)\\
\Omega^1(X)\otimes\mathbb{C} &= \Omega^{1,0}(X) \oplus \Omega^{0,1}(X)\\
\Omega^2(X)\otimes\mathbb{C} &= \Omega^{1,1}(X)
\end{align*}
and $\Omega^{p,q}(X) = 0$ for all other $p$ and $q$. Now note that
$$\Omega^{0,0}(X) \xrightarrow{\overline{\partial}} \Omega^{0,1}(X) \xrightarrow{\overline{\partial}} \Omega^{0,2}(X) = 0.$$
So $\overline{\partial}^2 = 0$ on $\Omega^{0,0}(X)$. Likewise, $\overline{\partial}^2 = 0$ on $\Omega^{1,0}(X)$, $\Omega^{0,1}(X)$, and $\Omega^{1,1}(X)$.
Added Later: Altenatively, we can show that for any almost complex structure $J$ on a smooth two-dimensional manifold, $N_J = 0$ where $N_J$ is the Nijenhuis tensor field of $J$. By the Newlander-Nirenberg Theorem, the vanishing of $N_J$ is equivalent to integrability.
Let $M$ be a smooth manifold with almost complex structure $J$. Recall that the Nijenhuis tensor field of $J$ is given by
$$N_J(X, Y) = [X, Y] + J[JX, Y] + J[X, JY] - [JX, JY].$$
As $N_J$ is tensorial, if $\{V_1, \dots, V_n\}$ is an ordered basis of local vector fields defined in a neighbourhood of $p$, $N_J$ vanishes at $p$ if and only if $N_J(V_i, V_j) = 0$ for all $1 \leq i, j \leq n$. Furthermore, $N_J$ is skew-symmetric, so it is enough to show that $N_J(V_i, V_j) = 0$ for all $1 \leq i < j \leq n$.
Now consider the case where $M$ is a smooth two-dimensional manifold. Fix $p \in M$ and let $V$ be a nowhere zero local vector field defined in a neighbourhood of $p$, then $\{V, JV\}$ is an ordered basis of local vector fields; to see this, note that if $JV \in \operatorname{span}\{V\}$ then $JV = kV$ so $-V = J^2V = k^2V$ which is impossible. As
\begin{align*}
N_J(V, JV) &= [V, JV] + J[JV, JV] + J[V, J^2V] - [JV, J^2V]\\
&= [V, JV] + J[V, -V] - [JV, -V]\\
&= [V, JV] - J[V, V] + [JV, V]\\
&= [V, JV] - [V, JV]\\
&= 0,
\end{align*}
we see that $N_J = 0$ at $p$. As $p$ is arbitrary, the Nijenhuis tensor field of $J$ vanishes, so $J$ is integrable.
Best Answer
To address the problem in the title, rather than the problem you set up in the subsequent text: let $\mathcal{X}_1$ and $\mathcal{X}_2$ denote two complex structures on an underlying manifold $X$. There are corresponding bundle maps $J_1,J_2:TX\to TX$. Suppose that $J_1=J_2$, or (more generally) that $f^*J_1=J_2$ for some diffeomorphism $f:X\to X$. Then we must show that $\mathcal{X}_1\cong\mathcal{X}_2$, as complex manifolds. This amounts to checking that $f$ (which in the simplest case is $f=\text{id}$) is a biholomorphism. Now prove the following:
Lemma. Let $f:\mathcal{X}\to\mathcal{Y}$ be a smooth map between complex manifolds with induced almost complex structures $J_X$ and $J_Y$. Then $f$ is holomorphic if and only if $df\circ J_X=J_Y\circ df$.
Using the lemma, you can now show that the assumption that $f^*J_1=J_2$ implies that $f$ must be a biholomorphism, and therefore $\mathcal{X}_1\cong\mathcal{X}_2$.