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.)
$X$ being locally compact and $\sigma$-compact metric implies that the one-point compactification $Y$ of $X$ is metrizable. Note that local compactness is needed to form the one-point compactification and that for $Y$ being metrizabale we need $\sigma$-compactness (since $\infty$ must have a countable basis of open neigborhoods; their complements are compact).
Note that all functions in $C_0(X)$ are bounded, thus $C_0(X)$ can be endowed with the $\sup$-norm.
Now let $i : X \to Y$ denote the inclusion map. Then
$$i^* : C(Y) \to C(X), i^*(f) = f \circ i = f \mid_ X$$
is a well-defined (linear) function.
If $C'(Y) = \{ f \in C(Y) \mid f(\infty) = 0 \}$, then $i^*(C'(Y)) = C_0(X)$. To see this, note that if $f \in C'(Y)$, then for each $\varepsilon > 0$ there exist an open neighborhood $U$ of $\infty$ in $Y$ such that $\lvert f(y) \rvert < \varepsilon$ for all $y \in U$. But $U = Y \setminus C$ for some compact $C \subset X$ which implies that $i^*(f)$ vanishes at infinity. Conversely, if $g \in C_0(X)$, define $e(g) : Y \to \mathbb R$ by $e(g)(x) = g(x)$ for $x \in X$ and $e(g)(\infty) = 0$. Continuity of $e(g)$ in all $x \in X$ is obvious because $X$ is open in $Y$. To check continuity in $\infty$, note that for each $\varepsilon > 0$ there exist a comapct $C \subset X$ such that $\lvert g(x) \rvert < \varepsilon$ for all $x \in X \setminus C$. Hence $\lvert e(g)(y) \rvert < \varepsilon$ for all $y \in Y \setminus C$.
Clearly, if $f, f' \in C'(Y)$ such that $i^*(f) = i^*(f')$, then $f = f'$. Hence we obtain a (linear) bijection
$$i^* : C'(Y) \to C_0(X) .$$
By construction it is an isometry. Since $C'(Y)$ is separable (as a subset of separable space), also $C_0(X)$ is separable.
Best Answer
Question 1: The Alexandroff compactification of a metrizable space is metrizable if and only if the original space is separable. Compact metrisable spaces are easily seen to be separable via the following lemma.
In particular, if $E^*$ is metrisable, $E$ is homeomorphically embedded in a separable space and hence is separable.
The converse follows from Urysohn's metrization theorem. It's obvious that $E^*$ is regular Hausdorff since it is compact and Hausdorff. It is a bit more fiddly but not very difficult to show that it is second countable by using the fact that $E$ itself must be second countable since it is a separable metric space.
However the converse obviously fails if $E$ is not metrisable since subspaces of metrisable spaces are metrisable.
Question 2: Part of the construction of $E^*$ includes that $E$ is homeomorphically embedded in to $E^*$ via some map $i$. It is then immediate that $f$ is càdlàg as a function valued in $E^*$ since the topological definition of limits is suitably compatible with the definition of the subspace topology.
For example, you want to check that left limits of $f$ exist in $E^*$ and are the same as the limits in $E$.
This means you want to show that for every $t \in (0,\infty)$ and every open neighbourhood $O^*$ of $\lim_{s \uparrow t} f(s)$ in $E^*$ (where the limit is taken in $E$) there is an $\varepsilon >0$ such that $f(s-\varepsilon,s) \subset O^*$. But $O^* \cap E$ is an open neighbourhood of $\lim_{s \uparrow t} f(s)$ in $E$ and so there is an $\varepsilon$ such that $f(s-\varepsilon,s) \subset O^* \cap E$ since is cadlag as an $E$-valued function.
The right-continuity is then similar.