$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then
$$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$
therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then
$f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to  
$u_x=v_y$, $u_y=-v_x$.

To prove the converse, make the linear change of the variables:
$z=x+iy,\; \overline{z}=x-iy$. A simple calculation shows that
$f_z=(1/2)(f_x-if_y),\; f_{\overline{z}}=(1/2)(f_x+if_y).$
Then the Cauchy-Riemann conditions in the new variables become $f_{\overline{z}}=0$.
Assuming that this is satisfied, set $h=f_z$. Then $$h_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$
so $h$ does not depend on $\overline{z}$ that is $h\in C[[z]]$.
Taking an antiderivative of $h$ in $C[[z]$ we obtain the $g$ as above.

Remark. This is how old textbooks like Whittaker Watson explain the Cauchy-Riemann conditions. Pure algebra.