There is a little bit of confusion regarding your question. For instance, you are assuming that $A$ is infinite countable, and from the statement in the title of your question, that could simply not be the case.
Here is a proof of the right hand side implication. First, since $A$ is infinite, then there exists an infinite countable subset of $A$ (this requires at least assuming AC$_\omega$). Let this set be denoted by $B$, and choose an enumeration of $B$; $B=\{b_n|\;n\in\omega\}$.
On the one hand, we clearly have that $A\preccurlyeq A\cup\{b\}$, via the injective function $i:A\longrightarrow A$ given by: for all $a\in A$, $\;i(a)=a$, that is, $i$ is the inclusion of $A$ into $A\cup\{b\}$.
On the other hand, the function $f:A\cup\{b\}\longrightarrow A$ defined by: for all $x\in A\cup\{b\}$:
$$f(x)=\begin{cases}
b_0 \qquad\text{if }x=b\\
b_{n+1}\quad\text{if }x=b_n\text{ for some }n\in\omega\\
x\qquad\text{ in any other case}
\end{cases}$$
Is clearly injective, so $A\cup\{b\}\preccurlyeq A$. From the Cantor-Bernstein theorem, we can conclude that $A\approx A\cup\{b\}$.
To prove the other implication, take into account that $A$ is a proper subset of the set $A\cup\{b\}$ which is equinumerous to $A$, so we immediately get that $A\cup\{b\}$ is infinite, as well as $A$ (in some literature this result is known as Dedekind's theorem; if you are iterested about this, i.e., that a set is infinite if and only if it contains a proper subset to which it is equinumerous, you should check my answer to this question).