= Solution
Use steady or initial-rate conditions with effectively clamped <Substrate>[substrate] <Concentration>[concentrations]. For either <Substrate>[substrate] $X\in\{S,R\}$, put $U_X=[AX]$ and $V_X=[AX^*]$. The intermediate balances are
$$
0=k_+^X[A][X]-(k_-^X+m)U_X,\qquad
0=mU_X-(l_-^{X*}+V_A)V_X.
$$
The <Reaction rate>[product flux] is therefore
$$
J_X=V_AV_X=
\frac{k_+^X[A][X]\,mV_A}{(k_-^X+m)(l_-^{X*}+V_A)}.
$$
The <Enzyme>[free enzyme] <concentration> is common to the two competing pathways, so it cancels even when <enzyme> sequestration changes its magnitude. Consequently
$$
\boxed{F=\frac{J_R}{J_S}=
\frac{[R]}{[S]}\frac{k_+^R}{k_+^S}
\frac{k_-^S+m}{k_-^R+m}
\frac{l_-^{S*}+V_A}{l_-^{R*}+V_A}.}
$$
Each post-binding factor is the ratio of <Probability>[probabilities] of proceeding rather than dissociating at that stage. The expression assumes positive common forward rates and no significant product back reaction. Without the steady/intermediate-balance or initial-rate assumption, the ratio can depend on intermediate populations and observation time; binding and <Dissociation rate constant>[dissociation] constants alone would not determine a unique transient ratio.
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