Recall $$ F =( dA_{ij} + A_{il}\wedge A_{lj} )\mu_i \otimes \mu_j^\ast $$ Hence if rank of $E$ is $2$, then $$ F= dA $$ since $A$ is skewsymmetric. If $D$ is Yang Mill connection then $ D^\ast F=0$. In addition, for $x\in \Omega^1({\rm Ad}\ E)$, $$ (dA , Dx) = (d^\ast dA, x) $$ Hence we have $$ d^\ast d A =0 $$
Here we want to prove that $$ d^\ast A=0$$
Proof : (1) In addition, for $x\in \Omega^0({\rm Ad}\ E)$ where ${\rm rank}\ E=2$, $$ (A,D x) = (A, dx + [A,x]) = (A,dx)=(d^\ast A,x) $$
(2) In textbook "If we require in addition to $d^\ast dA=0$ the gauge condition, $d^\ast A=0$" I can not understand this sentence.
(3) If gauge $s=e^u$ where $u(x_0)=0,\ x_0\in M$, then $$ s^\ast A = s^{-1} ds + s^{-1} As= du + A $$
(4) If $D$ is Yang Mill then $s^\ast D$ is so.
I guess that we can prove by using these facts. But I can not finish. Thank you in anticipation.