Call the "interesting" point in the Hawaiian earring $q$ (the point where all the circles intersect). In your space $X$, glue the quotiented point $\{H \times \{0\}\}$ to the point $(q,1) \in H \times I$, essentially bending your cone around in a loop. Call the resulting quotient space $Y$, and the glued point $p$. Note that $p$ could be described either as $(q,1)$ or as $\{H \times \{0\}\}$. ($Y$ can also be described as the Mapping Torus of the map that takes the entire Hawaiian earring to the point $q$).
$Y$ is semi-locally simply connected but doesn't satisfy property $∗$.
To see $Y$ is semi-locally simply connected, note that any point has a neighborhood that is disjoint from a set of the form $H \times \{x\}$ for some $x \in I$. The inclusion of such a neighborhood into $Y$ is nullhomotopic (simply retract the portion contained in $H \times [0,x)$ to $p$, then follow the strong deformation retraction of $X$ to $\{H \times \{0\}\}$), which certainly implies that any loop contained in it is nullhomotopic in $Y$.
But $p$ itself has no simply-connected neighborhood. Any such neighborhood $N$ contains a loop $L \times \{1\}$ in $H \times \{1\}$, which could only be contracted in $N$ by including all of $L \times I$ in $N$. In particular, you'd have to include all of $\{q\} \times I$ in $N$, But $\{q\} \times I$ is itself a loop which isn't nullhomotopic in $Y$, and so also isn't nullhomotopic in $N$. Thus, $N$ cannot be simply connected.
Pick $x\in E$ and consider an evenly covered neighbourhood $U$ of $p(e)$. Let $U_a$ be the copy of $U$ which $x$ belongs to. Then pick $g \neq 1$ in $\Delta(p)$: if $g(z)=u$ for $z,u \in U_a$ then $p(z)=p\circ g(z)=p(u) \implies z=u$.
Note that the thesis still holds if you remove the homeomorphism hypothesis.
Best Answer
You need to assume more about the action than just freeness; otherwise this statement is not true. For example, $\mathbb R$ acts freely on itself by translation, and $\mathbb R$ is simply connected, but $\mathbb R/\mathbb R$ is a single point, and its fundamental group is not isomorphic to $\mathbb R$.
In order to get the conclusion you want, you also have to assume that the action satisfies the following condition:
(Note that this implies, in particular, that $G$ acts freely.) Some authors call an action satisfying this condition a "properly discontinuous action"; but I don't like that term because different authors use it with inequivalent definitions, and also because it leads to oxymoronic expressions like "a continuous properly discontinuous action." Allen Hatcher in his book Algebraic Topology introduced the term covering space action for a continuous group action satisfying the condition above, and I've adopted that term in my books.
The basic fact is that if the action of $G$ on $X$ is a covering space action, then the quotient map $X\to X/G$ is a covering map. If in addition $X$ is simply connected, then $\pi_1(X/G)$ is isomorphic to $G$.
For more details, see this MSE answer; and for even more, see Chapters 11 and 12 in my book Introduction to Topological Manifolds.