Hypothesis: Let $G \ne H$ denote two non-trivial groups.
Goal: Show that $G * H$ has a trivial center (hence is non-abelian) and contains an element of infinite order.
Is my attempted proof below valid?
Attempt:
Let $g \in G$ and $h \in H$ s.t. $g$ and $h$ are not equal to the identity elements of $G$ and $H$, respectively. We can choose $g$ and $h$ this way since $G$ and $H$ are stipulated to be non-trivial.
Then $e \ne ghg^{-1}h^{-1} \in G*H$.
Indeed, if $[g,h]$ denotes $ghg^{-1}h^{-1}$, then
$$ e \ne [g,h]^i \ne [g,h]^j \ne e $$
for all naturals $i \ne j$.
Since $[g,h] \in G*H$ and $[g,h]^i \ne e$ for all $i \in \mathbb{N}$, we have that $[g,h]$ has infinite order in $G*H$.
Yet this means that $g$ and $h$ do not commute with each other in $G*H$ (when $g$ and $h$ are viewed as distinct elements in $G*H$).
Repeating the argument above for arbitrary elements $x \ne y$ in $G*H$ yields that $[x,y] \ne e$ in $G*H$ either (for all non-trivial elements $x,y \in G*H$).
Then it follows that any two non-trivial elements $x \ne y$ in $G*H$ do not commute with each other (since $[x,y] \ne e$ iff $x$ and $y$ do not commute with each other -- a fact from group theory).
Then $G*H$ has trivial center.
Then $G*H$ is in particular non-abelian.
And from (4), we have that $G*H$ has elements of infinite order. This completes the proof.
EDIT: Since (6) in the above proof is false, here is another attempt to show that the center of $G*H$ is trivial:
Let $x \in G*H$ s.t. $x$ is not equal to the empty word (i.e., $x \ne e$).
If it happens that $x \in G$ or $x \in H$ (i.e., $x = g_1$ or $x=h_1$ for some $g_1 \in G$ or $h_1 \in H$), then let $y$ be a non-trivial element in the opposing group. For example, if $x \in G$, then let $e_H \ne y \in H$ so that we have $[x,y] \ne e$ in $G*H$. The analogous result holds in case $x \in H$.
Now if it happens that $x \notin G$ or $x \notin H$, then we have that $x$ can be expressed in terms of its fully reduced form
$$ x = g_1 h_1 \cdot \ldots \cdot g_n h_n $$
provided that $g_1$ and $h_n$ may be the empty word.
Now if we let $y$ be the first non-empty word in the above expression (i.e., if we let $y = g_1$ or $h_1$ depending on whether $g_1$ is empty or not), then we will have that
$$ [x, y] \ne e $$
in $G*H$.
Then we have shown that in either case there exists some element $y$ in $G*H$ for which $[x,y] \ne e$. From group theory, we know that this is equivalent to saying that $x$ and $y$ don't commute with each other. Yet $x$ was arbitrary in $G*H$, and hence we have shown that any arbitrary (non-trivial) element in $G*H$ has another element which it does not commute with.
Then the center of $G*H$ is trivial.