Let $\tilde a,\tilde b:I\rightarrow X$ be lifts of $a,b:I\rightarrow Y$ satisfying $p\circ \tilde a=a$, $p\circ \tilde b=b$, and let $F_t:a\simeq_{\partial I}b:I\rightarrow Y$ be a relative homotopy satisfying $F(s,0)=a(s)$, $F(s,1)=b(s)$, $F(0,t)=a(0)=b(0)$ and $F(1,t)=a(1)=b(1)$. Now apply the HELP lemma in the diagram
$\require{AMScd}$
\begin{CD}
0\times I\cup I\times \partial I@>G'>> X\\
@Vi V V @VVpV\\
I\times I @>F>> Y
\end{CD}$\require{AMScd}$
where $G'(s,0)=\tilde a(s)$, $G'(s,1)=\tilde b(s)$, $G'(0,t)=\tilde a(0)=\tilde b(0)$. Note that $p\circ G'=F\circ i$. From the data the HELP lemma gives a map $G:I\times I\rightarrow X$ satisfying
$$G(s,0)=\tilde a (s),\quad G(s,1)=\tilde b(s),\quad G(0,t)=\tilde a(0)=\tilde b(0),\quad p\circ G(s,t)=F(s,t).$$
In particular $p(G(s,1))=p(\tilde a(1))=p(\tilde b(1))=F(s,1)=a(0)=b(0)$, so the assignment $s\mapsto G(s,1)$ is a path in $X$ contained in the fibre over $a(0)=b(0)$, going from $\tilde a(0)$ and ending at $\tilde b(0)$. Since the covering projection has discrete fibres such a path must be constant.
Hence $\tilde a(1)=\tilde b(1)$, and $G:\tilde a\simeq_{\partial I}\tilde b$ is a relative homotopy.
Other basepoints than $e_0$ are irrelevant. Here are two problems in your proof.
You say that $G = p^{-1} \circ H$ is a homotopy. But there is no map $p^{-1} : B \to E$ unless you know that $p$ is a homeomorphism. That is what you want to show, so you cannot use it here.
You say that $p_*$ injective implies $p$ injective. This is not true. $p_*$ is always injective, but no covering map which is not a homeomorphism is injective.
Edited: Your recent proof is essentially correct, but I think it should be more detailed. Here is a suggestion:
You start with a path $\gamma : [0,1] \to E$ such that $p(0) = e_0, p(1) = e_1$. Then $p \circ \gamma$ is a loop in $B$ beginning and ending at $b_0$. Since $B$ is simply connected, there exists a homotopy $H : [0,1] \times [0,1] \to B$ such that $H(x,0) = (p \circ \gamma)(x)$, $H(x,1) = b_0$ for all $x$ and $H(i,t) = b_0$ for $i=0,1$ and all $t$. Thus with $R = [0,1] \times \{1\} \cup \{0,1\} \times [0,1]$ we have $H(R) = \{b_0\}$.
There exists a unique lift $\overline{H} : [0,1] \times [0,1] \to E$ such that $\overline{H}(x,0) = \gamma(x)$ for all $x$. Then $\overline{H}(R) \subset p^{-1}(b_0)$. The fiber $p^{-1}(b_0)$ is discrete and $R$ is connected (in fact, $R \approx [0,1]$), therefore $\overline{H}(R) = \{e\}$ for some $e \in p^{-1}(b_0)$. But now $e_i = \gamma(i) = \overline{H}(i,0) = e$, i.e. $e_0 = e_1$.
Best Answer
Show that $G$ is a path homotopy between $\gamma$ and the constant loop $e_C$.
This should imply that $[\gamma] = [e_C]$ which means $\ker (p_*)$ is trivial, which means that $p_*$ is injective.