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.)
[For convenience, I will pretend in this answer that any compactification actually contains $X$ as a subspace, so I don't have to constantly be writing down the embedding maps.]
No. For instance, you can define a compactification $Y=X\cup\{\infty\}$ where the only neighborhood of $\infty$ is the entire space (and every open subset of $X$ remains open). There will not exist any morphism from this compactification to $X^*$ unless the topology on $X^*$ happens to be the same as the topology on $Y$ (i.e., the only compact closed subset of $X$ is the empty set; given your assumption that $X$ is KC and noncompact, this is impossible!).
A separate issue is that $X$ is open in $X^*$, so if you have some other compactification in which $X$ is not open, you cannot expect it to have a morphism to $X^*$. There are also uniqueness issues--a morphism to $X^*$ does not need to send all the new points to $\infty$ (for instance, if $X$ is uncountable with the cocountable topology, you could let $X'$ be $X$ together with one more point with the cocountable topology and let $Y$ be the 1-point compactification of $X'$, and then the new point of $X'$ can map to anywhere in $X^*$ and the map will still be continuous). With non-Hausdorff spaces, a continuous map is not determined by its values on a dense subset, so generally it will be very hard to get any sort of uniqueness property like this without stronger hypotheses.
If you restrict your definition of "compactification" to require $Y$ to also be KC and that $X$ is open in $Y$, then it is true that $X^*$ is the terminal compactification (assuming $X^*$ is a compactification at all by this definition--it won't always be KC). These hypotheses make it trivial to check that the map $Y\to X^*$ sending every new point to $\infty$ is continuous (the hypothesis that $X$ is open in $Y$ gives continuity at points of $X$, and the hypothesis that $Y$ is KC gives continuity at new points).
For uniqueness, suppose $h:Y\to X^*$ is a morphism of compactifications, and let $A=X\cup h^{-1}(\{\infty\})\subseteq Y$. Then I claim $A$ is compact. To prove this, note that $h^{-1}(\{\infty\})$ is closed in $Y$ and hence compact, so it suffices to show any ultrafilter $F$ on $X$ has a limit in $A$. By compactness of $Y$, $F$ has a limit $y\in Y$; if $y\in A$ we're done, so we may assume $y\not\in A$. In that case $h(y)\neq\infty$, so it is a point of $X$, and then since $h$ is the identity on $X$, $F$ must converge to $h(y)$ in $X$. Thus $h(y)$ is a limit of $F$ in $A$.
Thus since $Y$ is KC, $A$ is closed in $Y$. Since $A$ contains $X$ and $X$ is dense in $Y$, this means $A=Y$. Thus $h$ must map every point of $Y\setminus X$ to $\infty$.
Best Answer
Update: Proofs adapted for one-point compactifications, which are not necessarily Hausdorff.
Let $X$ be a Hausdorff, first countable. Let $X^* = X \cup \{\infty\}$ be its one-point compactification. Let $(x_n)_{n \in \mathbb N}$ be a sequence in $X$.
$(x_n)_{n \in \mathbb N}$ converges to $\infty$, iff $(x_n)_{n \in \mathbb N}$ has no in $X$ convergent subsequence.
"$\Rightarrow$" is true even without first countability: Let $(x_n)_{n \in \mathbb N}$ converge to $\infty$. Assume $(x_{n_k})_k$ converges to $x \in X$. Then $K: = \{x_{n_k}: k \in \mathbb N\} \cup \{x\} \subset X$ is compact, hence closed in $X$, since $X$ is Hausdorff. Hence $(X \setminus K) \cup \{\infty\}$ is a neighborhood of $\infty$, thus $(x_{n_k})_k$ does not converge to $\infty$. Contradiction.
"$\Leftarrow$": Assume the sequence $(x_n)_{n \in \mathbb N}$ in $X$ does not converge to $\infty$. Then there is an open neighborhood $U$ of $\infty$ and a subsequence $(x_{n_k})_{k}$ such that $x_{n_k} \notin U$ for all $k$. Hence $A := \overline{\{x_{n_k}: k \in \mathbb N\}}$ does not contain $\infty$ (although the closure to be taken in $X^\star$). Hence, $A$ is compact and first countable, therefore sequentially compact (compare the above comment from Jephph). Thus, $(x_{n_k})_{k}$ contains a in $A \subset X$ convergent subsequence, which is also a subsequence of $(x_n)_n$.
Although this was not asked, it might be worthwhile to note that "$\Leftarrow$" does not hold without first countability: Let $x \in \beta \mathbb N \setminus \mathbb N$, $X := \beta \mathbb N \setminus \{x\}$. Then $X$ is locally compact, Hausdorff, not compact. Its one-point-compactification is homeomorphic to $\beta \mathbb N$. The sequence $(n)_{n \in \mathbb N}$ has no convergent subsequence neither in $X$ nor in $\beta \mathbb N$, since $\beta \mathbb N$ has no non-trivial convergent sequences at all. In particular, it does not converge to "$\infty$".