You can do it directly using the unit-counit zig-zag identities. What you are trying to do is the argument for 1. implies 2. in the following lemma with $K=\mathrm{id}_{\mathcal A}$ (sorry, the names of my objects differ from yours).
Lemma. Consider $1$-morphisms $\mathcal A\xrightarrow{F}\mathcal B$, $\mathcal A\xleftarrow{G}\mathcal B$, and a $2$-morphism $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF$. Then the following conditions are equivalent
$\DeclareMathOperator{\Lan}{Lan}$
There exists a $2$-morphism $FG\overset\epsilon\Rightarrow \mathrm{id}_{\mathcal B}$ satisfying the zig-zag identities
- $F\overset{F\eta}\Rightarrow FGF\overset{\epsilon F}\Rightarrow F$ is the identity $2$-morphism $F\overset{\mathrm{id}_F}\Rightarrow F$
- $G\overset{\eta G}\Rightarrow GFG\overset{G\epsilon}\Rightarrow G$ is the identity $2$-morphism $G\overset{\mathrm{id}_G}\Rightarrow G$
$G=\Lan_F^\eta\mathrm{id}_{\mathcal A}$ is an absolute left Kan extension, i.e. such that $KG=\Lan_F^{K\eta}K$ for any $1$-morphism $\mathcal A\xrightarrow{K}\mathcal C$.
$\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF$ is a $2$-morphism such that
The left extension of $\mathcal A\xrightarrow{F}\mathcal B$ along itself given by $F\overset{F\eta}\Rightarrow FGF$ is weakly initial among such extensions, i.e. has a morphism of extensions to any other extension.
$\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF$, considered as a left extension of $\mathcal A\xrightarrow{\mathrm{id}_{\mathcal A}}\mathcal A$ along $\mathcal A\xrightarrow{F}\mathcal B$, has no non-identity endomorphisms
Proof of 1. implies 2..
Consider a left extension of $\mathcal A\xrightarrow{K}\mathcal C$ along $\mathcal A\xrightarrow{F}\mathcal B$ given by a $2$-morphism $K\overset\tau\Rightarrow HF$. We claim that the unique morphism of extensions $KG\overset{\phi}\Rightarrow H$ is given by $KG\overset{\tau G}\Rightarrow HFG\overset{H\epsilon}\Rightarrow H$.
To show that the above defines a factorization, observe that the $2$-moprhism $KG\overset{\tau G}\Rightarrow HFG\overset{H\epsilon}\Rightarrow H$ is in fact an extension in two steps. First, $K\overset{K\eta}\Rightarrow KGF\overset{\tau GF}\Rightarrow HFGF\overset{H\epsilon F}\Rightarrow HF$ is, by naturality/horizontal composition/interchange law, the same as $K\overset{\tau}\Rightarrow HF\overset{HF\eta}\Rightarrow HFGF\overset{H\epsilon F}\Rightarrow HF$. Second, use the first zig-zag identity to conclude that this is precisely $K\overset\tau\Rightarrow HF$.
To show uniqueness of the factorization, suppose $K\overset\tau\Rightarrow HF$ factors as $K\overset{K\eta}\Rightarrow KGF\overset{\phi F}\Rightarrow HF$ for a $2$-morphism $KG\overset\phi\Rightarrow H$. We again proceed in two steps. First, the composite $KG\overset{K\eta G}\Rightarrow KGFG\overset{\phi FG}\Rightarrow HFG\overset{H\epsilon}\Rightarrow G$ is, by naturality/horizotnal composition/interchange law, the same as $KG\overset{K\eta G}\Rightarrow KGFG\overset{KG\epsilon}\Rightarrow KG\overset{\phi}\Rightarrow H$. Second, the latter is the same as $KG\overset{K\mathrm{id}_G}\Rightarrow KG\overset{\phi}\Rightarrow H$ by th second zig-zag identity.
Proof of 3. implies 1. Begin by observing that the first zig-zag identity says precisely that $F\overset{\mathrm{id}_F}\Rightarrow F$, interpreted as an extension $\mathrm{id}_{\mathcal B}F\Rightarrow F$ of $\mathcal A\xrightarrow{F}\mathcal A$ along itself, has a morphism of extensions $FG\overset\epsilon\Rightarrow\mathrm{id_{\mathcal B}}$ from the extension $F\overset{F\eta}\Rightarrow FGF$. But the existence of this morphism is precisely the weakly initial hypothesis on the extension $F\overset{F\eta}\Rightarrow FGF$.
Next, observe that, given the first zig-zag identity, the $2$-endomorphism
$G\overset{\eta G}\Rightarrow GFG\overset{G\epsilon}\Rightarrow G$ is in fact an endomorphism of the left extension $\mathrm{id_{\mathcal A}}\overset\eta\Rightarrow GF$. Indeed, the composite $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF\overset{\eta GF}\Rightarrow GFGF\overset{G\epsilon F}\Rightarrow GF$ is by naturality/horizontal composition/interchange law the same as $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF\overset{GF\eta}\Rightarrow GFGF\overset{G\epsilon F}\Rightarrow GF$, and the first zig-zag identity says that this is precisely $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF\overset{\mathrm{id}_GF}\Rightarrow GF$. But by hypothesis, $\mathrm{id}_{\mathcal A}\overset\eta\Rightarrow GF$ has no non-identity endomorphisms, hence the second zig-zag identity holds.
Corollary. Because the zig-zag identities are $1$- and $2$-dual, the following conditions are equivalent:
$\DeclareMathOperator{\Lift}{Lift}
\DeclareMathOperator{\Rift}{Rift}
\DeclareMathOperator{\Ran}{Ran}$
- $F\dashv G$ with unit $\mathrm{id}_A\overset\eta\Rightarrow GF$ and counit $FG\overset\epsilon\Rightarrow\mathrm{id}_A$
- $G=\Rift_\epsilon^F\mathrm{id}_B$ is an absolute right Kan lift
- $G=\Rift_\epsilon^F\mathrm{id}_B$ and $GF=\Rift_{\epsilon F}^FF$ are right Kan lifts
- $F=\DeclareMathOperator{\Lift}{Lift}\Lift_\eta^G\mathrm{id}_A$ is an absolute left Kan lift
- $F=\Lift_\eta^G\mathrm{id}_A$ and $FG=\Lift_{\eta G}^G G$ are left Kan lifts
- $G=\Lan^\eta_F\mathrm{id}_A$ is an absolute left Kan extension
- $G=\Lan^\eta_F\mathrm{id}_A$ and $FG=\Lan^{F\eta}_FF$ are left Kan extensions
- $F=\Ran^\epsilon_G\mathrm{id}_B$ is an absolute right Kan extension
- $F=\Ran^\epsilon_G\mathrm{id}_B$ and $GF=\Ran^{G\epsilon}_GG$ are right Kan extensions
