Given an adjunction $F\dashv G:\mathcal{C}\rightleftarrows\mathcal{D}$ with unit $\eta$ and counit $\epsilon$, we naturally have a monad $(G\circ F,\eta,G\epsilon_F)$ on $\mathcal{C}$ and a comparison functor $K:\mathcal{D}\to\mathcal{C}^{G\circ F}$ (where $\mathcal{C}^{G\circ F}$ is the E-M category of this monad) given by $$K(D)=\big(G(D),G(\epsilon_D)\big),$$ $$K(f:D\to D')=G(f).$$ This functor is unique satisfying $$?\circ K=G, \hspace{10mm} K\circ F=\widehat {G\circ F},$$ as shown in e.g. Mac Lane, p.142.

It seems like the proof that this functor is unique satisfying these equations should be way easier than it is in Mac Lane. He uses the fact that $$(1_\mathcal{C},K):F\dashv G:\mathcal{C}\rightleftarrows\mathcal{D}\longrightarrow\widehat{G\circ F}\dashv\ ?:\mathcal{C}\rightleftarrows\mathcal{C}^{G\circ F}$$ is a morphism of adjunctions by the above equations and the fact that both adjunctions have the same unit, then looks at the equivalent counit condition for morphisms of adjunctions to conclude that any other functor $K':\mathcal{D}\to\mathcal{C}^{G\circ F}$ satisfying $?\circ K'=G$ and $K'\circ F=\widehat{G\circ F}$ agrees with $K$ on structure maps.

Why is this not immediately true since $K$ and $K'$ agree on arrows?

In particular, any two functors $F,G:\mathcal{A}\rightrightarrows\mathcal{B}$ which agree on arrows immediately agree on objects since $$F(X)=F(dom(1_X))=dom(F(1_X))=dom(G(1_X))=G(dom(1_X))=G(X),$$ and the equation $?\circ K=G=\ ?\circ K'$ tells us that $K$ and $K'$ agree on arrows -- they're both just $G$ on arrows since $?$ leaves arrows unchanged.

Can we just repeat the above argument with $F=K$ and $G=K'$ and be done?

I can't find anything wrong with this reasoning, but I suspect Mac Lane would have taken this route if it worked.