Suppose $X$ is a CW-complex of dimension $n$. If $e_i$ is an $n$-cell then is $H_n(X) \to H_n(X, X \setminus e_i)$ the zero map? Suppose $X$ is a CW-complex of dimension $n$. If $e_i$ is an $n$-cell then is $H_n(X) \to H_n(X, X \setminus e_i)$ the zero map? If not, then is $H_n(X, X\setminus e_i) \to H_n(X, X \setminus \{x\})$ the zero map where $x$ is a point in the interior of $e_i$?
I cannot prove either, I suspect the proof would boil down to some fact about the homology of CW complexes which I do not know.
For what its worth, these statements are relevant for a proof I am reading where $X$ is a compact manifold of dimension $n$ admitting a CW structure and homology with $\mathbb{Z}_2$ coefficients are taken.
 A: No, take $X = S^n = e_0 \cup e_n$ as a counterexample. The map $H_n(X) \to H_n(X, X \setminus \{x\})$ is an isomorphism.
More generally your map $X \to X/(X \setminus e_n) \cong S^n$ is a cellular map. It induces a chain map $f: C_*^{CW}(X, *) \to C_*^{CW}(S^n, *) = \Bbb Z[n]$, where the latter notation means we have a copy of the integers in degrees 0 and n and no differential. The map $f$ collapses every generator except $e_n$.
If $c = \sum c(j) e_n^j$ is a sum of n-cells with let's say $e_n^0 = e_n$ is the fixed one in the discussed above, then $f(c) = c(0)$. If $f$ is zero on homology then every cycle has $c(0) = 0$; if $f$ is nonzero on homology there is some cycle with $c(0) \neq 0$.
Putting this all together your map is zero on homology if and only if $\partial e_n$ is not a sum of boundaries of other cells. I'm not sure if this is what you're looking for, it's more or less just from definition.
A: I have an answer that doesn't require you to know much about the homology of CW complexes but requires a bit more work. We're going to look at the long exact sequences for the pairs you mention.
First, we have a simple example to use for your first question that might illuminate the general case. Let $X = S^2$ with CW structure given by one $0$-cell, one $1$-cell (with constant attaching map), and two $2$-cells $e_1$ and $e_2$ that give each of the hemispheres. Then $X \smallsetminus e_1$ is a (open) hemisphere (i.e., the interior of $e_2$), and $H_2(X, X\smallsetminus e_1)$ by excision is isomorphic to $H_2(e_2/\partial e_2)$ which is free of rank $1$. This means that the long exact sequence for the pair $(X, X \smallsetminus e_1)$ is
$\cdots \to H_2(X \smallsetminus e_1) \to H_2(X) \to H_2(X, X \smallsetminus e_1) \to H_1(X \smallsetminus e_1) \to \cdots$
Now, $X \smallsetminus e_1$ is contractible, so exactness tells us that the map $H_2(X) \to H_2(X, X\smallsetminus e_1)$ is an isomorphism of free rank-$1$ abelian groups. So, it's not the zero map.
What about your second question? A strategy for figuring this one out for general $X$ can be to consider the long exact sequences for $(X, X \smallsetminus e_i)$ and $(X, X \smallsetminus *)$ (I'll be using $*$ to denote $\{ x\}$; it's a little easier to type.), and use naturality of the sequences with respect to the map of pairs $f: (X, X\smallsetminus e_i) \to (X, X\smallsetminus *)$ induced by the identity, and perhaps naturality will tell us how $f_*$ behaves.
In this case, we have two long exact sequences connected by $f_*$, the relevant portion of which is
$\require{AMScd}
\begin{CD}
H_n(X \smallsetminus e_i) @>>> H_n(X) @>>> H_n(X, X \smallsetminus e_i) @>>> H_{n-1}(X \smallsetminus e_i)  \\
 @Vf_*VV & @V1VV @V{f_*}VV @VVf_*V &  \\
H_n(X \smallsetminus *) @>>> H_n(X) @>>> H_n(X, X \smallsetminus *) @>>> H_{n-1}(X \smallsetminus *) \end{CD}$
Now, consider the situation where $X$ has a single $n$-cell; this means that $X \smallsetminus e_i$ and $X \smallsetminus *$ deformation retract to the $(n-1)$-skeleton of $X$, so the two groups on the left vanish since $H_n$ of an $(n-1)$-dimensional CW complex is always zero. Exactness tells us that the maps $H_n(X) \to H_n(X, X\smallsetminus e_i)$ and $H_n(X) \to H_n(X, X\smallsetminus *)$ are at least injective, and commutativity of the center square implies that $f_*$ cannot be the zero map. If we use the example $X = S^n$ with usual CW structure (a single zero-cell and a single $n$-cell), all the groups on the left and right vanish and we end up with a commutative square where all maps are immediately seen to be isomorphisms besides $f_*$, but from that it follows that $f_*$ is an isomorphism.
This is perhaps a bit more complicated of an answer than it needs to be (see the comment that explains why $f_*$ is always an isomorphism), but hopefully this approach is also helpful to see. In the situation where you have a compact manifold and are using $\mathbb{Z}_2$ coefficients, it is always possible to have a CW structure where you have a single $n$-cell, so  we're always in the above situation; if you want to use integer coefficients you just need to add the hypothesis that $X$ is oriented.
