This excercise is quite long an involved, and I think you'll learn a lot from going through the details yourself. So, this answer will thus be purposely incomplete.
Let's once and for all fix an element $\tilde{e}\in X$ with the property that $p(\tilde{e}) = e\in G$, where $e$ denotes the identity in $G$.
Our first goal is to create a map $\tilde{\mu}:X\times X\rightarrow X$. As hinted in the comments, we already have a map $\mu\circ (p\times p):X\times X\rightarrow G$, so if we can lift that map, we have our desired $\tilde{\mu}$. Such a lift exists if and only if $$\mu_\ast((p\times p)_\ast(\pi_1(X\times X, (\tilde{e},\tilde{e}))\subseteq p_\ast(\pi_1(X,\tilde{e})),$$ so let's check this condition. We need a lemma.
Lemma 1: The induced map $\mu_\ast: \pi_1(G,e)\times \pi_1(G,e)\rightarrow \pi_1(G,e)$ is $\mu_\ast([\gamma],[\alpha]) = [\gamma][\alpha]$ where juxtaposition on the right denotes the group operation in $\pi_1(G,e)$.
Proof: Consider the composition $G\xrightarrow{i_1} G\times G\xrightarrow{\mu} G$ where $i_1(g) = (g,e)$. This composition is the identity on $G$. Using $1$ to denote the trivial loop, it follows that $\mu_\ast([\gamma], 1) = [\gamma]$ for all $[\gamma]\in\pi_1(G,e)$.
Repeating this using $i_2(g) = (e,g)$, we find that $\mu_\ast(1,[\gamma]) = [\gamma]$ for all $[\gamma]\in \pi_1(G,e)$.
Finally, any element in $\pi_1(G,e)\times \pi_1(G,e)$ has the form $([\gamma],[\alpha]) = ([\gamma],1)(1,[\alpha])$ for some $[\gamma],[\alpha]\in pi_1(G,e)$. Since $\mu_\ast$ is a homomorphism, we therefore have $$\mu_\ast([\gamma],[\alpha]) = \mu_\ast([\gamma],1)\mu_\ast(1,[\alpha]) = [\gamma][\alpha],$$ as claimed. $\square$
With the lemma in hand, we can now prove:
Proposition 2: We have $\mu_\ast( (p\times p)_\ast(\pi_1(X\times ,(\tilde{e}, \tilde{e}))))\subseteq p_\ast(\pi_1(X,\tilde{e})).$
Proof: For ease of writing, I'm going to write $H:=p_\ast(\pi_1(X,\tilde{e})).$ Then our goal is to show that $\mu_\ast(H\times H)\subseteq H$. But, from Lemma 1, $\mu_\ast(H\times H) = HH\subseteq H$ since $H$ is a subgroup. $\square$
So, we have a lift $\tilde{\mu}:X\times X\rightarrow X$. Notice that $\tilde{\mu}(\tilde{e},\tilde{e})$ projects to $e$. Thus, we may pick $\tilde{\mu}$ so that $\tilde{\mu}(\tilde{e},\tilde{e}) =\tilde{e}$. Once we have made this choice, $\tilde{\mu}$ is unique. Here is the first important fact regarding $\tilde{\mu}$.
Proposition 3: The projection map $p$ resepcts $\tilde{\mu}$ in the sense that $p\circ \tilde{\mu}(x_1,x_2) = \mu(p(x_1), p(x_2))$. In other words, $p$ is a homomorphism (except we don't yet know that $\tilde{\mu}$ gives a group structure.)
Proof: This is exactly what is meant by saying that $\tilde{\mu}$ is a lift of $\mu\circ (p\times p). $\square$
But how can we verify that $\tilde{\mu}$ is a group multiplication? There are three things we need to verify: associativity, the existence of an identity, and the existence of inverses.
Here's the argument for inverses (which I believe is the hardest case). Something similar works for the other two properties which need verification.
First, we need an inverse map. To that end, consider the composition $X\xrightarrow{p}G\xrightarrow{inv} G$ where $inv(g) = g^{-1}$ is the inverse map on $G$. I will leave it to you to verify that this composition has a unique lift $\tilde{inv}:X\rightarrow X$ with $\widetilde{inv}(\tilde{e}) = \tilde{e}$.
But why is $\widetilde{inv}$ an inverse map? We need to verify that $\tilde{\mu}(x,\widetilde{inv}(x)) = \tilde{e}$ for all $x\in X$. Said another way, we need to verify that the composition $X\xrightarrow{(i_1, \widetilde{inv})}X\times X\xrightarrow{\tilde{\mu}} X$ the constant map with image $\tilde{e}$.
To that end, consider the composition $X\xrightarrow{p}G\xrightarrow{c_{e}} G$ where $c_e$ denotes the map which is constantly $e$: $c_e(g) = e$ for all $g\in G$. I claim that the map $\tilde{\mu}\circ(i_1,\widetilde{inv})$ lifts this map. Indeed, we have $$p(\tilde{\mu}((i_1,\widetilde{inv})(x)) = \mu(p\times p)(x, \widetilde{inv}(x)) = \mu(p(x), p(\widetilde{inv}) = \mu(p(x), inv(p(x)) = \mu(p(x), p(x)^{-1}) = e.$$
So, the map $\tilde{\mu}\circ(i_1, \widetilde{inv})$ and the map $c_{\tilde{e}}:X\rightarrow X$ are both lifts of the same map agreeing at one point, so they agree everywhere. That is, $\widetilde{inv}$ really is the inverse map.
I will skip the tedious proof of well-definition and group axioms since you have already got them. Hence I will only focus on the proof of the continuity of the map
$$\star\colon E\times E\to E, \qquad (a,b)\mapsto a\star b^{-1}$$
From the construction and the fact that $p$ is a group homomorphism, we have a natural commutative diagram
$$\require{AMScd}
\begin{CD}
E\times E @>{\star}>> E\\
@VVV @VVV \\
G\times G @>{\cdot}>> G
\end{CD}$$
where the vertical lines are the product map $p\times p$ and $p\colon E\to G$ respectively, and the horizontal arrow is the continuous map $\cdot \colon G\times G\to G$ sending $(g_1,g_2)\mapsto g_1\cdot g_2^{-1}$.
$\textbf{Definition}$ Given an open subset $U\subseteq E$, I say that $U$ is ${\it elementary}$ with respect to $p$ if $p_{|U} \colon U\to p(U)$ is an homeomorphism.
$\textbf{Lemma:}$ It there exists a topological basis of $E$ given by elementary sets with respect to $p$.
$\textit{proof:}$ Let $U$ be an open set of $E$ and $x\in U$. It there exists an open neighbourhood $V$ of $p(x)$ such that $p^{-1}(V)=\sqcup_{\alpha}U_\alpha$ and $p_{|U_\alpha}\colon U_\alpha\to V$ is an homeomorphism. Let $U_x:=U_{\alpha(x)}$ be the unique open elementary set of the family containing the point $x$. Then $U_x\cap U\subset U$ is elementary with respect to $p$. By the arbitrary choice of $x\in U$, we get $U=\cup_{x\in U}U_x\cap U$.
$\textbf{Remark:}$ The definition of a open elementary set together with the above lemma works for any covering $p\colon X\to Y$.
$\textbf{Proposition:}$ The above map $\star$ is continuos.
$\textit{proof:}$ For simplicity of notation, let us define $f:=\cdot \circ (p\times p)$. From the previous lemma, it is sufficient to prove the continuity for any open elementary set $U$ of $E$ with respect $p$. Hence we only need to prove that $\star^{-1}(U)$ is open. Given $x\in \star^{-1}(U)$, since $E\times E$ is locally path connected, then it there exists a open path connected neighbourhood $V$ of $x$. Moreover, this can be chosen such that $f(V)\subseteq p(U)$ (remember that $p(U)$ is open in $G$). Given a point $y=(y_1,y_2)\in V$, we choose a path $\gamma$ from $(e, e)$ to $x$ and a path $\eta$ from $x$ to $y$ all contained in $V$. Then $
f_*(\gamma\eta)$ is a path from $1_G$ to $p(y_1)p(y_2^{-1})$, which is homotopic to the path $p(f(t))p(g(t))$ that you defined above. Thus their lifting via $p$ starting at $e$ are homotopic and in particular have the same ending point $\star(y)$. By construction, since $U$ is elementary, we observe that the lifting of $
f_*(\gamma\eta)$ is simply $p^{-1}\circ f(\gamma\eta)$, so that the end point $\star(y)$ falls into $U$. Hence $y\in \star^{-1}(U)$ and from the arbitrary choice of $y\in V$ we have $V\subset \star^{-1}(U)$. This proves $\star^{-1}(U)$ is open.
Please let me know if you agree with me or if you have some doubts and questions :).
Best Answer
Consider the maps $\alpha,\beta\colon\widetilde G^3\longrightarrow\widetilde G$ defined by$$\alpha(g_1,g_2,g_3)=\widetilde\mu\left(\widetilde\mu(g_1,g_2),g_3\right)\quad\text{and}\quad\beta(g_1,g_2,g_3)=\widetilde\mu\left(g_1,\widetilde\mu(g_2,g_3)\right).$$You want to prove that they are equal. Note that, if $g_1,g_2,g_3\in\widetilde G$, then\begin{align}p\left(\alpha(g_1,g_2,g_3)\right)&=p\left(\widetilde\mu\left(\widetilde\mu(g_1,g_2),g_3\right)\right)\\&=p\left(\widetilde\mu(g_1,g_2)\right)\cdot p(g_3)\\&=p(g_1)\cdot p(g_2)\cdot p(g_3)\\&=p(g_1)\cdot p\left(\widetilde\mu(g_2,g_3)\right)\\&=p\left(\widetilde\mu\left(g_1,\widetilde\mu(g_2,g_3)\right)\right)\\&=p\left(\beta(g_1,g_2,g_3)\right).\end{align}In other words, $p\circ\alpha=p\circ\beta$. Since, furthermore, $\alpha\left(\widetilde e,\widetilde e,\widetilde e\right)=\beta\left(\widetilde e,\widetilde e,\widetilde e\right)=\widetilde e$, $\alpha=\beta$.