You get the uniqueness result if the space is Hausdorff.
Let $\langle X,\tau\rangle$ be a compact space. Suppose that $p\in X$ is in the closure of $Y=X\setminus\{p\}$, and let $\tau_Y$ be the associated subspace topology on $Y$; $\langle X,\tau\rangle$ is then a compactification of $\langle Y,\tau_Y\rangle$.
Suppose that $p\in U\in\tau$, and let $V=U\cap Y$. Then $\varnothing\ne V\in\tau_Y$, so $Y\setminus V$ is closed in $Y$. Moreover, $Y\setminus V=X\setminus U$ is also closed in $X$, which is compact, so $Y\setminus V$ is compact. That is, every open nbhd of $p$ in $X$ is the complement of a compact, closed subset of $Y$. Thus, if $\tau'$ is the topology on $X$ that makes it a copy of the Alexandroff compactification of $Y$, then $\tau\subseteq\tau'$.
Now let $K\subseteq Y$ be compact and closed in $Y$, and let $V=Y\setminus K\in\tau_Y$. If $X\setminus K=V\cup\{p\}\notin\tau$, then $p\in\operatorname{cl}_XK$. If $X$ is Hausdorff, this is impossible: in that case $K$ is a compact subset of the Hausdorff space $X$ and is therefore closed in $X$. Thus, if $X$ is Hausdorff we must have $\tau=\tau'$, and $X$ is (homeomorphic to) the Alexandroff compactification of $Y$.
If $X$ is not Hausdorff, however, we can have $\tau\subsetneqq\tau'$. A simple example is the sequence with two limits. Let $D$ be a countably infinite set, let $p$ and $q$ be distinct points not in $D$, and let $X=D\cup\{p,q\}$. Points of $D$ are isolated. Basic open nbhds of $p$ are the sets of the form $\{p\}\cup(D\setminus F)$ for finite $F\subseteq D$, and basic open nbhds of $q$ are the sets of the form $\{q\}\cup(D\setminus F)$ for finite $F\subseteq D$. Let $Y=D\cup\{q\}$. Then $Y$ is dense in $X$, and $X$ is compact, and $Y$ itself is a closed, compact subset of $Y$ whose complement is not open in $X$.
Improved example (1 June 2015): Let $D$ and $E$ be disjoint countably infinite sets, let $p$ and $q$ be distinct points not in $D\cup E$, let $X=D\cup E\cup\{p,q\}$, and let $Y=D\cup E\cup\{q\}$. Points of $D\cup E$ are isolated. Basic open nbhds of $q$ are the sets of the form $\{q\}\cup (E\setminus F)$ for finite $F\subseteq E$, and basic open nbhds of $p$ are the sets of the form $\{p\}\cup\big((D\cup E)\setminus F\big)$ for finite $F\subseteq D\cup E$. Then $Y$ is a non-compact dense subspace of the compact space $X$, so $X$ is a (non-Hausdorff) compactification of $Y$. Let $K=\{q\}\cup E$. Then $K$ is a compact closed subset of $Y$, but $X\setminus K=\{p\}\cup D$ is not open in $X$.
(This avoids the question of whether it’s legitimate to look at the Alexandrov compactification of a compact space.)
If $X$ is not compact, then every one-point compactification of $X$ is an open embedding. Indeed, suppose $X$ is not open in a one-point compactification $X^*=X\cup\{\infty\}$. This means there is some $x\in X$ such that every open neighborhood of $x$ in $X^*$ contains $\infty$. But then every open neighborhood of $x$ contains a neighborhood of $\infty$, and thus its complement is a compact subset of $X$. Since $X$ is embedded in $X^*$, this means that every open neighborhood of $x$ in $X$ has compact complement. But this implies $X$ is compact, since every open cover of $X$ includes a set which contains $x$, and then the complement of that set is covered by finitely many other sets in the cover.
Best Answer
We need to also assume that $A$ is not open. Let $x\in X\setminus A$, then since $X$ is compact Hausdorff, and $x\notin A$, there is a neighbourhood $U$ of $x$ whose closure is disjoint from $A$ and is compact (follows from the regularity of $X$). So, $X\setminus A$ is locally compact, Haudorff.
The restriction of the quotient map $q\colon X\to X/A$ to $X\setminus A$ is an injective continuous map whose image misses the point corresponding to $A$. Moreover, the closure of this image is $X/A$ because $A$ is not open (else $X\setminus A$ is already compact).
Being the continuous image of a compact set, $X/A$ is compact and by regularity of $X$, it is Hausdorff. By the uniqueness of one point compactification, $X/A$ is the one point compactification of $X\setminus A$.