First of all we shall reduce the RHS fraction, so $ \bbox[lightyellow] { {1 \over a} + {1 \over b} = {p \over {10^{\,n} }} = {{p/\gcd (p,10^{\,n} )} \over {10^{\,n} /\gcd (p,10^{\,n} )}} = {q \over A} }$
Then we have that $ \bbox[lightyellow] { \eqalign{ & {A \over a} + {A \over b} = q\quad \Rightarrow \cr & \Rightarrow \quad \left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor + \left\{ {{A \over a}} \right\} + \left\{ {{A \over b}} \right\} = q\quad \Rightarrow \cr & \Rightarrow \quad \left[ \matrix{ \left( {\left\{ {{A \over a}} \right\} = 0} \right)\; \wedge \;\left( {\left\{ {{A \over b}} \right\} = 0} \right)\; \wedge \;\left( {\;\left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor = q} \right)\quad \quad (1) \hfill \cr \quad \quad \vee \hfill \cr \left( {\left\{ {{A \over a}} \right\} + \left\{ {{A \over b}} \right\} = 1} \right)\; \wedge \;\left( {\;\left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor = q - 1} \right)\quad \quad \quad \;\;(2) \hfill \cr} \right. \cr} }$ where $x = \left\lfloor x \right\rfloor + \left\{ x \right\}$ is the decomposition into floor and fractional part
Now the first condition gives $ \bbox[lightyellow] { \left( {c\backslash A} \right)\; \wedge \;\left( {d\backslash A} \right)\; \wedge \;\left( {\;c + d = q} \right)\quad \Rightarrow \quad \left\{ \matrix{ a = A/c \hfill \cr b = A/d \hfill \cr} \right .\quad\quad (1) }$ i.e., if there are divisors of $A$ summing to $q$, then we can get $a$ and $b$ as thei quotient of $A$ with such divisors.
The second condition translates to $ \bbox[lightyellow] { \left( {{{A\bmod a} \over a} + {{A\bmod b} \over b} = 1} \right)\; \wedge \;\left( {\;\left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor = q - 1} \right)\quad \quad \quad \;\;(2) }$ and there is not much more to manage.