I reject the premise of the question. :-)
It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.
The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next.
Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.
Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )
The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.
The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □
Compare the complexity of this argument to the other arguments supplied so far.
Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.
Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.
The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.
Ahhhh, suppose $\psi_g^k(1) = 1$, i.e. that $g\Phi(g)\cdots\Phi^{k-1}(g) = 1$. Then $g = \psi^{k+1}_g(g) = g\Phi(g)\cdots \Phi^{k-1}(g)\Phi^k(g) = \Phi^k(g)$, so $g$ is $\Phi$-periodic with period dividing $k$. Therefore if there is a uniform bound to the period of $\Phi$-periodic elements, then there is a uniform bound (depending on $\Phi$) for the period of principal elements for $\Phi$.
This latter statement holds for $F$ a finite-rank free group, say of rank $n$. I claim the subgroup $P(\Phi)$ of $F$ comprising all $\Phi$-periodic elements has rank at most $n$. (Since the restriction of $\Phi$ to $P(\Phi)$ is periodic, this will show there is a bound on the period.) Indeed, suppose $x_0,\ldots,x_n$ are $n+1$ elements of a free basis for $P(\Phi)$. There exists $N$ such that each $x_i$ has period dividing $N$, thus they belong to the fixed subgroup of $\Phi^N$, which has rank at most $n$ by Bestvina–Handel's Theorem (the Scott Conjecture), a contradiction.
Indeed, the argument in the first paragraph shows that the result holds for a group $F$ provided that there is a bound on the period of $\Phi$-periodic elements.
Best Answer
Yes, and even with an inner automorphism.
Namely, start from your example of a countable group $G$ with elements $x_n$ ($n$ ranging over some arbitrary subset) and automorphism $f$ such that $x_n$ lies in an $n$-cycle for $f$. Let $G'$ be the semidirect product $G\rtimes\langle f\rangle$. Finally, embed $G'$ into a finitely generated group $H$ (every countable group embeds into a f.g. group, an old result of Higman-Neumann-Neumann). Then in $H$, for the inner automorphism defined by $f$, the element $x_n$ lies in an $n$-cycle. [NB: as $G$ can be chosen to be abelian and hence $G'$ metabelian, $H$ can be chosen to be 4-step solvable, by the Neumann-Neumann proof of the previous embedding theorem.]
Note: there are no examples among linear groups (= subgroup of $\mathrm{GL}_d$ over a field) with an inner automorphism. Indeed, for the inner automorphism $i_g$ defined by an element $g$, the centralizer of $g^{n!}$ is an increasing sequence of subgroups. But in a matrix group the centralizers in the algebra of matrices are subspaces. So there is no strictly infinite sequence of centralizers. Hence the finite $i_g$-cycles have bounded length.
Here's another example, due to B.H. Neumann 1937, with the additional feature of being residually finite. Namely, consider a set $X$ consisting of the infinite disjoint union of sets $X_n$ ($n$ ranging over an infinite set $I$ of integers $\ge 5$), each $X_n$ being in bijection with $\{1,\dots,n\}$. Let $c$ be the element acting as the $n$ cycle $n\mapsto 1\mapsto 2\mapsto\dots$ on each $X_n$, and $t$ the element acting as the transposition $1\mapsto 2\mapsto 1$ on each $X_n$. Let $G_I$ be the subgroup of permutations of $X$ generated by $\{t,c\}$. Then $G_I$ is residually finite (since it acts faithfully with finite orbits). Also $G_I$ contains, for each $n\in I$, as normal subgroup, the alternating group $\mathrm{Alt}_n$ of $X_n$ (acting trivially elsewhere). Let $s_n$ be the 3-cycle $1\mapsto 2\mapsto 3\mapsto 1$ acting on $X_n$, identity elsewhere. Then we see that $s_n\in G_I$, and the inner automorphism defined by $c$ admits $s_n$ as element of period exactly $n$.