I'm not completely sure if this proof works that an isomorphism preserves the order of elements. Here is my attempt at a proof.
Let $f: G \to H$ be an isomorphism of groups and $x \in G$. Then $$\begin{align} x^n = e_G &\iff f(x^n) = f(e_G) \\ &\iff f(x)^n = e_H. \end{align}$$ The second biconditional is true because $f$ is injective. The first implication implies that the order of $f(x)$ divides the order of $x$. The backwards implication implies that the order of $x$ divides the order of $f(x)$. So $\text{order}(x) =\text{order}(f(x))$.
The reason I'm not fully convinced is because most of the proofs of this fact I've read do not use a single line of biconditionals, but invoke the fact that if $|a| = n$ and $a^k = e$, then $n \mid k$. I haven't explicitly done that, so I worry my answer makes a leap of logic.
Apart from the typos mentioned in the comments, the proof attempt is missing a lot of needed detail.
For example, what is $n$ when you write $x^n=e_G$. You do not say if it is the order of $x$ in $G$. It could be simply $0$, a negative integer, or any integer multiple of $\operatorname{ord}_G(x)$. In fact, a proper proof attempt will elucidate on why, in the first place, if $x^n=e_G$ for some integer $n$, it must be an integer multiple of $\operatorname{ord}_G(x)$
The definition of $\operatorname{ord}_G(x)$ is not just any integer such that $x^n=e_G$ but rather, the least positive integer, if it exists, such that $x^n=e_G$. You will need to make use of this fact, that if say $\operatorname{ord}_G(x)=n$, then $x^n=e_G$ but $x^k\neq e_G$ for all $1\le k\lt n$
Also, the presented attempt does not take into account the cases when $\operatorname{ord}_G(x)=\infty$, i.e., does not exist.