Dynamic accumulator with only non-membership witness
This is not a correct dynamic accumulator scheme.

Your scheme assumes that the following holds: if $a,b$ are primes and  $0&lt; a,b &lt; p$, then $$(a,b)=1 \implies (ab\mod p,a)\neq1. $$ If this is true, then your scheme is correct, however this is not the case. 

Toy example: $p=11,a=5,b=7$. Since $ab \mod 11=35 \mod 11=2,$ therefore $A=2$. Now, it is clear that security does not hold, since one could prove false non-membership statements. The reason for that is: $(a,A)=1=(b,A)$. Meaning one can prove that $a$ and $b$ are not accumulated however they were indeed added to the accumulator. For instance $(2,1)$ is a a non-membership witness for $5$, when $A=2$, since $5*2+2*1 \mod 11 = 1$.