For 1, yes you have an $E$ around $\infty$ with $E$ homeomorphic to a ball, but you don't know anything about $\overline{E}$. In fact, $\overline{E}$ is potentially all of $M^\ast$!
On the other hand, inside of $E$ is another open set $E'$, also homeomorphic to a ball, but whose closure is homeomorphic to $\overline{\mathbb{B}^n}$. Use $U = M^\ast \setminus \overline{E'}$.
For 2, since $M\setminus U$ is homeomorphic to $\mathbb{R}^n\setminus \mathbb{B}^n$, it follows that $M\setminus\overline{U}$ is homeomorphic to $\mathbb{R}^n\setminus \overline{\mathbb{B}^n}$. Call such a homeomorphism $g$.
By using the inversion map $f$, one sees that $\mathbb{R}^n\setminus\overline{\mathbb{B}^n}$ is homeomorphic to $\mathbb{B}^n \setminus\{\vec{0}\}$.
Composing $g$ and $f$, we have a homeomorphism between $M\setminus\overline{U}$ and $\mathbb{B}^n\setminus \{\vec{0}\}$. Try to prove that we can use these to find a homeomorphism between $M^\ast \setminus \overline{U}$ and $\mathbb{B}^n$.
[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
We first show that
$$\sigma(\tau_{X^*})=\sigma(\tau_X\cup\{\infty\})=:\mathscr{F}$$
as you claimed. Let $U\in \tau_{X^*}$ so that $U\in \tau_X$ or $U=\{\infty\}\cup V$ where $V$ is the complement of a compact set in the space $(X,\tau_X)$. In the former case, we immediately have $U\in\mathscr{F}$. In the latter, we note that every compact set in $(X,\tau_X)$ is closed (by the Hausdorff property), so $V,V\cup\{\infty\}\in\mathscr{F}$. This proves $\sigma(\tau_{X^*})\subseteq \mathscr{F}$. Conversely, note that $\tau_X\subseteq\tau_{X^*}$ and in particular $X\in\tau_{X^*}$ so that $\{\infty\}=X^*\setminus X\in\sigma(\tau_{X^*})$. This completes the proof of the claim.
Note that the collection $$\mathscr{G}:= \sigma(\tau_X) \cup \{U\cup\{\infty\} : U\in\sigma(\tau_X)\}$$ is a $\sigma$-algebra. Since $\mathscr{G}$ contains every set in $\tau_X\cup\{\infty\}$, it contains $\mathscr{F}$. Conversely, $\mathscr{F}$ contains $\sigma(\tau_X)$ along with all sets of the form $U\cup\{\infty\}$ for $U\in\sigma(\tau_X)$. Thus $\mathscr{G}=\mathscr{F}$ as required.