A generalization of holomorphic functions Assume that $U$ is an open set in the complex plane $\mathbb{C}$ and $A$ is a real $2\times 2$ matrix.
We define
$$\mathcal{S}_{A}=\{f:U\to \mathbb{C}\mid Df.A=A.Df  \}$$
where $Df$ is the $2\times 2$ Jacobian matrix of smooth  function $f:U\to \mathbb{R}^{2}\simeq \mathbb{C}$.
Every function with this property is called an $A$-holomorphic function.
Example: When $J=\begin{pmatrix} 0&-1\\1&0\end{pmatrix}$, $\mathcal{S}_{J}$ is the space of (usual)  holomorphic functions.

Assume that this vector space  $\mathcal{S}_{A}$ is closed under "uniform convergence on compact subsets of $U$". That is: Assume that $f_{n}$ is a sequence in $\mathcal{S}_{A}$ and $f_{n}$ converges to $f$, uniformly on compact  subsets of  $U$. Then $f$ must  be in $\mathcal{S}_{A}$.
Does this imply that $A$ is in the form $$A=\begin{pmatrix} a&-b\\b&a \end{pmatrix}$$

The question is  motivated by the fact that the uniform limit of  a sequence of  holomorphic  functions is  a holomorphic  function.
 A: Simple counterexample: 
Suppose $f(x,y) = u(x,y) + i v(x,y)$ is a holomorphic function. Then it satisfies the Cauchy-Riemann equations
\begin{align}
u_x = v_y\\
v_x = -u_y 
\end{align}
Then the function $g(x,y) = \tilde{u}(x,y) + i \tilde{v}(x,y)$ given by 
\begin{align}
\tilde{u}(x,y) = \frac12 u(2x,y) \\
\tilde{v}(x,y) = v(2x,y) 
\end{align}
satisfies
\begin{align}
\tilde{u}_x = \tilde{v}_y \\
\tilde{v}_x = -4 \tilde{u}_y 
\end{align}
which is the same as $ A.D g = D g.A $ for 
$$ A = \begin{pmatrix} 0 & -1/4 \\ 1 & 0 \end{pmatrix} $$

This example is obtained geometrically by doing a linear change of variables $(x,y) \mapsto (2x,y)$ on the domain and $(u,v) \mapsto (\frac12 u,v)$ on the co-domain and so all convergence and closure of holomorphic functions carry over unchanged. 
More generally, the same procedure shows that for any $A$ of the form 
$$ A = c P J P^{-1} $$
where $P\in GL(2)$ and $c\in \mathbb{R}_+$ the closure property asked for in the question holds. 

[Edit to answer this question]: 
The correct way to think about the question is that the two copies of $A$ in the equation $A (Df) = (Df) A$ are not the same. One is the complex structure on the domain and the other is the complex structure on the codomain. If you conjugate the complex structure on the domain side ONLY what you get is $$ P A P^{-1} (Df) = (Df) A$$ where the second $A$ is unchanged because it is the complex structure on the codomain. To make the equation work as written, we then do the reverse conjugation on the complex structure of the codomain which changes the equation to $$ PA P^{-1} (Df) = (Df) P A P^{-1}$$
The action on the codomain side is the reverse (given by the inverse matrix) simply because covariance versus contravariance in the equation. 
A: There is a counterexample:
Let $A=\begin{pmatrix} 0&1\\1&0\end{pmatrix}$,then $\mathcal{S}_{A}=\{f:U\to \mathbb{C}\mid \frac{d}{dx}f = \frac{d}{dy}f\}$.
$\mathcal{S}_{A}$ is closed, but $A$ is not like $\begin{pmatrix} a&-b\\b&a \end{pmatrix}$.
