This is a general fact about sequential colimits (or even more generally colimits over posets). $\newcommand\colim{\operatorname{colim}}$
Let $A : \mathbb{N}\to \mathcal{C}$ be some sequential diagram.
$\mathbb{N}$ has an endomorphism $s$ defined by $s(n)=n+1$.
There is a corresponding natural transformation $\sigma : A \to s^*A$,
which at $n$ is the map $f_n : A_n\to A_{n+1}$ which is $A(n\le n+1)$, the transition map in the sequential diagram.
In fact, more generally, let $t : \mathbb{N}\to \mathbb{N}$ be any nondecreasing map, then $t$ gives an endomorphism of $\mathbb{N}$ as a category, so we can consider the pullback $t^*A$. If $t$ additionally satisfies $t(n)\ge n$ for all $n$, then there is a natural transformation $\tau: A\to t^*A$ which at $n$ is the unique map $A_n\to A_{t(n)}$ in the diagram $A$.
Now if we assume that $t$ is cofinal (for all $n\in \mathbb{N}$, there exists $N\in\mathbb{N}$ such that $t(N)\ge n$), the following is true.
Proposition. $\colim A$ exists if and only if $t^*\colim A$ exists. In this case if $t(n)\ge n$ for all $n$, so $\tau$ exists, then $\tau$ induces a morphism $\tilde{\tau}:\colim A\to \colim t^* A$ on passing to colimits which is an isomorphism.
Proof.
It suffices to give a natural isomorphism between $\newcommand\cocone{\operatorname{co-cone}}\cocone_A$ and $\cocone_{t^*A}$, the functors that the colimits represent. Given some cocone to $A$, $(X, i_n: A_n\to X)$, we get a cocone to $t^*A$ by taking $(X, i_{t(n)}: A_{t(n)}\to X)$. However, it turns out that we can also go backwards, since $t$ is cofinal. If we are given $(Y,j_m : A_{t(m)}\to Y)$, we can define
$i_n : A_n\to Y$ by taking $N$ large enough that $t(N)\ge n$, and then defining $i_n$ to be the composite $A_n \to A_{t(N)} \xrightarrow{j_N} Y$. This is well defined regardless of the choice of $N$ because $\mathbb{N}$ is a poset and $(Y,j_m)$ are a cocone to $t^* A$. This gives us a cocone $(Y, i_n)$ to $A$. These operations are inverse to each other, so we have a natural isomorphism as desired.
Finally we need to show that when $t(n)\ge n$ for all $n$, and if $\colim A$ exists then $\tau$ induces an isomorphism on the colimits. First let's fix some notation. Let $i_n : A_n\to \colim A$ be the inclusions for the colimit of $A$, let $j_n : A_{t(n)}\to \colim t^*A$ be the inclusions of the colimit of $t^*A$, and let $\tau_n: A_n\to A_{t(n)}$ be the components of $\tau$.
Then we observe that $\tilde{\tau}$ is defined by
applying the universal property of the colimit $\colim A$ to the cocone
$(\colim t^*A, j_n \circ \tau_n)$. But we know (up to isomorphism) what the $j_n$ are. By the first half of this proof the $j_n$ are $i_{t(n)}$, so
$$j_n\circ \tau_n = i_{t(n)}\circ \tau_n = i_n,$$
since $\tau_n : A_n\to A_{t(n)}$ is a map in the diagram $A$ and $i$ is a
cocone to $A$. Thus after potentially composing with an isomorphism $(\colim t^*A, j_n)\to (\colim A, i_{t(n)})$, $\tilde{\tau}$ is actually given by the colimiting cone itself, so we have that for some isomorphism $\phi$, $\phi\circ \tilde{\tau} = 1_{\colim A}$, and thus $\tilde{\tau}$ is an isomorphism itself (namely $\phi^{-1}$). $\blacksquare$
Edited to fix the final argument
Build a modified version of the stable homotopy group colimit sequence as follows.
Fix $a$, let $P_{2n} = \pi_{n+a}(X_a)$, let $P_{2n+1} = \pi_{n+a}(\Omega\Sigma X_a)$, let the maps $f_{2n} : P_{2n}\to P_{2n+1}$ be $\pi_{n+a}(\eta_{X_a})$, and let the maps $f_{2n+1} : P_{2n+1}\to P_{2n+2}$ be given by the composite of $\pi_{1+n+a}$ applied to the structure map $\Sigma X_a\to X_{a+1}$ with the isomorphism $\pi_{n+a}(\Omega\Sigma X_a)\cong \pi_{n+1+a}(\Sigma X_a)$.
Let $d: \mathbb{N}\to \mathbb{N}$ be given by $d(n) =2n$, and let
$\delta : P\to d^*P$ be the corresponding natural transformation.
Note that $d^*P$ is precisely the sequence that usually defines the stable homotopy groups.
Now consider the $\sigma$ natural transformation, $\sigma_P$, for this sequence. If we restrict to even $n$, then $\sigma_P$ corresponds to $\pi_{\bullet+a}(\eta)$. In other words $d^*\sigma_P = \pi_{\bullet + a}(\eta)$. We have a commutative diagram
$$
\require{AMScd}
\begin{CD}
P @>\sigma_P>> s^* P\\
@V\delta VV @V\delta VV \\
d^*P @>>\pi_{\bullet+a}(\eta)> d^*s^*P.\\
\end{CD}
$$
Termwise this commutative diagram is
$$
\require{AMScd}
\begin{CD}
P_n @>>> P_{n+1}\\
@VVV @VVV \\
P_{2n}=\pi_{n+a}(X_a) @>>\pi_{n+a}(\eta_{X_a})> P_{2n+1}=\pi_{n+a}(\Omega \Sigma X_a).\\
\end{CD}
$$
Now if we take colimits of this diagram of functors, three of the maps, $\sigma_P$ and both $\delta$s induce isomorphisms on the colimit. Therefore the last map does as well.
Best Answer
The point is that $\eta\circ k=\eta+\eta+\dots+\eta$ is the $k$-fold sum of $\eta$ in $\pi_3S^2$. In particular $\eta\circ(-1)=-\eta$. Now, $$k\circ\eta=\eta\circ k^2$$ and $$\Sigma(k\circ\eta)=\Sigma k\circ\Sigma\eta=\Sigma\eta\circ\Sigma k=\Sigma(\eta\circ k)$$ by the commutativity of these elements after suspension. Hence $$\Sigma(\eta\circ k)=\Sigma(\eta\circ k^2).$$
With $k=-1$ we have $$\Sigma(\eta\circ(-1))=\Sigma(-\eta)=-\Sigma\eta$$ on the left, and $$\Sigma(\eta\circ(-1)^2)=\Sigma(\eta\circ 1)=\Sigma\eta$$ on the right. In particular $$\Sigma\eta=-\Sigma\eta,$$ showing that this element has order no greater than $2$.
Thus to show that $\Sigma\eta$ has order $2$ it will suffice to show that it is nontrivial. One way to accomplish this is by noting that the mapping cone construction commutes with suspension. In particular, the mapping cone of $\Sigma\eta$ is $\Sigma\mathbb{C}P^2$. If $\Sigma\eta$ were null-homotopic, then $\Sigma\mathbb{C}P^2$ would be homotopy equivalent to $S^3\vee S^5$ (which is the mapping cone of the constant map $S^4\rightarrow S^3$). That $\Sigma\mathbb{C}P^2$ does not split as a wedge of spheres is a consequence of the non-trivial Steenrod square $$Sq^2:H^3(\Sigma\mathbb{C}P^2;\mathbb{Z}/2)\rightarrow H^5(\Sigma\mathbb{C}P^2;\mathbb{Z}/2).$$
If this is unfamiliar, then let us stick to the EHP sequence. We know that the suspension $\Sigma:\pi_3S^2\rightarrow\pi_4S^3$ is surjective by Freudenthal's Theorem. Since $\eta$ generates $\pi_3S^2\cong\mathbb{Z}$, and $\Sigma\eta$ has order $\leq2$, the group $\pi_4S^3$ is either $\mathbb{Z}/2$ or is trivial. Both of these groups are devoid of odd torsion, so to answer our problem it will suffice to work 2-locally.
Thus introduce the EHP sequence, $$\dots\rightarrow\pi_5S^5\xrightarrow{P}\pi_3S^2\xrightarrow\Sigma\pi_4S^3\rightarrow 0$$ which becomes exact after localisation at $2$. By exactness, the right-hand group is trivial if and only if $P$ is surjective. But it is well-known that $$P(\iota_5)=[\iota_2,\iota_2],$$ where $[\iota_2,\iota_2]$ is the Whitehead product of $\iota_2$ with itself. I calculated the valued of this Whitehead product here (at least up to sign), showing that $$[\iota_2,\iota_2]=\pm2\eta$$ (the exact sign will depend on your particular conventions).
The point is that $P$ is not surjective, so it must be that $\pi_4S^3\cong\mathbb{Z}/2$, generated by $\Sigma\eta$.