= Solution
Absorb the gauge coupling into a matrix-valued <connection 1-form>, and use
$$
F=dA+A\wedge A,qquad
D\alpha=d\alpha+A\wedge\alpha-(-1)^p\alpha\wedge A
$$
for an adjoint-valued $p$-form $\alpha$. The <matrix trace> supplies the usual invariant nondegenerate pairing of the <Lie algebra>, and the spacetime metric defining the <Hodge star> is fixed during the gauge-field variation. Since $\delta F=D\delta A$ and the two-form Hodge pairing is symmetric,
$$
\begin{aligned}
\delta I
&=2\int_M\operatorname{tr}(D\delta A\wedge *F)\\
&=2\int_{\partial M}\operatorname{tr}(\delta A\wedge *F)
+2\int_M\operatorname{tr}(\delta A\wedge D*F).
\end{aligned}
$$
For variations fixed at the boundary, the <Yang-Mills equations> are therefore
$$
\boxed{D*F=0.}
$$
They accompany, rather than replace, the <gauge-theory Bianchi identity> $DF=0$, which follows directly by substituting $F=dA+A\wedge A$ and using $d^2=0$.
Fix the <Yang-Mills gauge transformation> convention
$$
A^g=g^{-1}Ag+g^{-1}dg,qquad F^g=g^{-1}Fg.
$$
Taking $g=1+\eta+O(\eta^2)$ gives
$$
\boxed{\delta_\eta A=d\eta+[A,\eta]=D\eta,qquad
\delta_\eta F=[F,\eta].}
$$
The curvature variation also follows from $D(D\eta)=[F,\eta]$. If the group parameter is defined with the opposite sign, both displayed infinitesimal variations change sign together.
The <Hodge star> acts only on spacetime indices, so $\delta(*F)=[*F,\eta]$. Consequently
$$
\begin{aligned}
\delta_\eta\operatorname{tr}(*F\wedge F)
&=\operatorname{tr}([*F,\eta]\wedge F+*F\wedge[F,\eta])\\
&=\operatorname{tr}([*F\wedge F,\eta])=0.
\end{aligned}
$$
This proves \b[exact off-shell <gauge invariance>, with no surviving boundary term], even for a gauge parameter nonzero at the boundary. It does not require the <Yang-Mills equations>.
One can check the surface terms explicitly in the integrated variation. The Lie-algebra-valued four-form $[F,*F]=F\wedge *F-*F\wedge F$ vanishes: writing $F=F^iT_i$, the coefficients $F^i\wedge *F^j$ are symmetric in $i,j$, and multiply the antisymmetric commutator $[T_i,T_j]$. Thus $D^2*F=[F,*F]=0$, and the gauge substitution in the variation gives
$$
\delta_\eta I=2\int_{\partial M}\operatorname{tr}(D\eta\wedge *F+\eta D*F)
=2\int_{\partial M}d\operatorname{tr}(\eta *F)=0.
$$
The last equality is <Stokes theorem> on the closed boundary; equivalently the cancellation is already pointwise in the preceding derivation. Keeping only the first surface term from an arbitrary variation would incorrectly produce a gauge boundary term.
For the additional <Yang-Mills theta term>, denote its coefficient-free integral by $P=\int_M\operatorname{tr}(F\wedge F)$. Its gauge variation is again pointwise zero:
$$
\delta_\eta\operatorname{tr}(F\wedge F)=\operatorname{tr}([F\wedge F,\eta])=0.
$$
Under an arbitrary connection variation, its <boundary variation of the Yang-Mills theta term> is
$$
\begin{aligned}
\delta P
&=2\int_M\operatorname{tr}(D\delta A\wedge F)\\
&=2\int_{\partial M}\operatorname{tr}(\delta A\wedge F)
+2\int_M\operatorname{tr}(\delta A\wedge DF)
=2\int_{\partial M}\operatorname{tr}(\delta A\wedge F).
\end{aligned}
$$
The bulk term vanishes by the <gauge-theory Bianchi identity>. Therefore \b[adding a constant multiple of $P$ does not change the bulk <Yang-Mills equations>]. With fixed boundary potential, it also leaves the usual variational problem unchanged. If boundary variations are allowed, it changes the boundary term: $I+\kappa P$ has surface variation $2\int_{\partial M}\operatorname{tr}(\delta A\wedge(*F+\kappa F))$, so compatible boundary conditions or a boundary action must account for that change.
Locally $\operatorname{tr}(F\wedge F)=d\operatorname{tr}(A\wedge dA+\tfrac23 A\wedge A\wedge A)$, the unnormalized <Chern-Simons 3-form> identity. That three-form need not be globally defined on a nontrivial bundle, so absence of a bulk variation does not force $P$ itself to vanish. Its exact <gauge invariance> is the curvature-conjugation calculation above, independently of the choice of local gauge potential.
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