Here are two definitions of a primary ideal.
An ideal $I\subset A$ is primary if $I\ne A$ and $xy\in I\implies$ either $x\in I$ or $y^n\in I$ for some $n> 0$.
An ideal $I\subset A$ is primary if $Ass(A/I)$ consists of a single element.
Are they equivalent?
$2.\implies1.$ Suppose $Ass(A/I)$ consists of a single element $P$. If $I=A$, then I believe $Ass(A/I)=Ass(\{0\})$. The zero ring has no prime ideals, so there are no associated primes of the element $0$ in the zero ring. This contradiction shows that $I\ne A$. But I don't know how to deduce the other condition.
I don't have ideas for the other implication either.
I assume $A$ is Noetherian.
$1\implies 2$. Suppose $I$ is a proper ideal such that if $xy\in I$, then either $x\in I$ or $y^n\in I$ for some $n>0$.
Then if $P$ is associated to $A/I$, there exists $x\not \in I$ in $A$ such that $P$ is the annihilator of $x$ in $A/I$. I.e., $Px\subseteq I$, but this implies that $$I\subseteq P\subseteq \sqrt{I},$$ so $\sqrt{I}=P$. Thus the only possible associated prime of $A/I$ is $\sqrt{I}$, and since $A/I$ is nonzero, it has an associated prime. Thus $\operatorname{Ass} A/I$ consists of a single element, $\sqrt{I}$.
$2\implies 1$. Suppose $\operatorname{Ass} A/I$ consists of a single element. Then if $xy\in I$, with $x\not\in I$, then $y$ annihilates the image of $x$ in $A/I$, so $y$ lies in some associated prime, necessarily the unique associated prime of $A/I$, $P$. Since every ideal minimal over $I$ is associated to $A/I$, $P$ is the unique minimal prime over $I$, and therefore $P=\sqrt{I}$. Thus since $y\in P$, $y^n\in I$ for some $n$.