I have come up with a counterexample:
$S^2$ and $S^3 \times \mathbb{C}P ^\infty$ have the same homotopy groups, but their suspensions don’t since $\pi_7 (S^3)$ is torsion but $\pi_7(S(S^3 \times \mathbb{C}P ^\infty))$ contains an infinite element since $\pi_7(S^4)$ is not torsion.
For any space $Y$ there is an isomorphism
$$\iota:H^*(Y\vee Y)\xrightarrow\cong H^*Y\oplus H^*Y,\qquad \gamma\mapsto (i_1^*\gamma,i_2^*\gamma),$$
where
$$Y\xrightarrow{i_1}Y\vee Y\xleftarrow{i_2}Y$$
are the inclusions into each factor. The inverse is given by
$$\kappa:H^*Y\oplus H^*Y\xrightarrow\cong H^*(Y\vee Y),\qquad (\alpha,\beta)\mapsto q_1^*\alpha+q_2^*\beta$$
where
$$Y\xleftarrow{q_1}Y\vee Y\xrightarrow{q_2}Y$$
are the maps collapsing opposite factors to a point.
Below we'll need to know that these isomorphisms are compatible with the action of any cohomology operation $\Phi$. This is easy to check for $\iota$, for which naturality provides
$$\iota(\Phi_{Y\vee Y}(\gamma))=(i_1^*\Phi_{Y\vee Y}(\gamma),i_2^*\Phi_{Y\vee Y}(\gamma))=(\Phi_Y(i_1^*\gamma),\Phi_Y(i_2^*\gamma))=\Phi_Y\oplus\Phi_Y(\iota(\gamma))$$
for any $\gamma\in H^*(Y\vee Y)$. For the inverse compute
$$\Phi_{Y\vee Y}\circ\kappa=\kappa\circ\iota\circ(\Phi_{Y\vee Y}\circ \kappa)=\kappa\circ(\Phi_{Y\vee Y}\oplus\Phi)\circ(\iota\circ\kappa)=\kappa\circ(\Phi_Y\oplus\Phi_Y).$$
Now, specialise to the case $Y=\Sigma X$ for some space $X$. Recall that the suspension is equipped with a comultiplication
$$c:\Sigma X\rightarrow \Sigma X\vee\Sigma X$$
satisfying
$$q_1\circ c\simeq id_{\Sigma X}\simeq q_2\circ c.$$
Consequently, if $\alpha,\beta\in H^*(\Sigma X)$, then
$$c^*(\kappa(\alpha,\beta))=\alpha+\beta.$$
Finally, suppose that $\Phi$ is a stable cohomology operation and $\alpha,\beta\in H^*X$. For notational reasons write $\alpha'=\sigma(\alpha),\beta'=\sigma(\beta)\in H^{*+1}(\Sigma X)$ for the suspensions of $\alpha,\beta$. Then
$$\Phi_X(\alpha+\beta)=\Phi_{\Sigma X}(\alpha'+\beta')=\Phi_{\Sigma X}(c^*\kappa(\alpha,\beta))=c^*(\Phi_{\Sigma X\vee\Sigma X}(\kappa(\alpha',\beta')))=c^*\kappa(\Phi_{\Sigma X}\alpha',\Phi_{\Sigma X}\beta')=\Phi_{\Sigma X}\alpha'+\Phi_{\Sigma X}\beta'=\Phi_X\alpha+\Phi_X\beta$$
which is what was to be shown.
Best Answer
The trick is here is to not start with the cofiber, but rather to find it naturally inside another space, where it sits in a manner in which we can easily understand it. Details are as follows.
Consider the infinite complex projective space $\mathbb{C}P^\infty$. Then the integral cohomology ring $H^*\mathbb{C}P^\infty\cong\mathbb{Z}[x]$ is a polynomial algebra on a degree $2$ class $x$. It follows that $\mathbb{C}P^\infty$ has a cell structure $S^2\cup e^4\cup\dots\cup e^{2n}\cup\dots$, and that the inclusion $S^2\cong\mathbb{C}P^1\hookrightarrow \mathbb{C}P^\infty$ induces an isomorphism on $H^2$. Moreover, after reduction mod $p$ we have $\mathcal{P}^1x=x^p\in H^{2p}(\mathbb{C}P^\infty;\mathbb{Z}_p)$ by the properties of the Steenrod squares.
The point is that after suspension we have a map $S^3\hookrightarrow\Sigma\mathbb{C}P^\infty$ which induces an isomorphism on $H^3$ with any coefficients and we know exactly how $\mathcal{P}^1$ acts on the domain of the induced homomorphism. As it stands it seems difficult to get much out of this, but it turns out that there is a lot we can do if we introduce an odd prime $p$ and consider the same problem after localisation at this prime. If you are unfamiliar with this process I will be happy to supply some references, but the main idea is that we are formally inverting the action of every other prime $p'\neq p$ as a multiplication operator on $\pi_*$ and $H_*$. It is a celebrated result that that this algebraic construction can actually be realised topologically.
Now $\mathbb{C}P^\infty$ is a homotopy-associative, homotopy-commutative H-space, since it identifies homotopically with both $BS^1$ and $K(\mathbb{Z},2)$. Let $\mu:\mathbb{C}P^\infty\times\mathbb{C}P^\infty\rightarrow\mathbb{C}P^\infty$ be its multiplication. In the sequel I will be following Cohen's paper Splitting Certain Suspensions Via Self-Maps.
At this stage we Introduce the odd prime $p$ and choose a mod $p$ primitive root of unity $k$. We use this to form the map $f_k$ as the composition
$$f_k:\mathbb{C}P^\infty\xrightarrow{\Delta}\prod^k\mathbb{C}P^\infty\xrightarrow{\mu(\mu\times1)\dots(\mu\times1\times\dots\times 1)}\mathbb{C}P^\infty$$
where the first map is the iterated diagonal. We know for dimensional reasons that in cohomology we have $\mu^*x=x\otimes 1+1\otimes x$ and an easy calculation shows us that
$$f^*_k(x^n)=k^n\cdot x^n.$$
Now with $k$ fixed as above we define for each $j=1,\dots,p-1$ a map $\theta_j:\Sigma\mathbb{C}P^\infty\rightarrow\Sigma\mathbb{C}P^\infty$ by
$$g_j=\Sigma f_k-k^j$$
where the sum is taken using the suspension comultiplciation of $\Sigma\mathbb{C}P^\infty$, and $k^j$ is the map which is $k^j$ times the identity (if you identify $\Sigma\mathbb{C}P^\infty=S^1\wedge \mathbb{C}P^\infty$, then this map is just the wedge $k^j\wedge 1$ of the degree map on $S^1$). Another easy check shows us that
$$g_j^*(\sigma x^n)=(k^n-k^j)\sigma x^n$$
where $\sigma x^n\in H^{2n+1}\Sigma\mathbb{C}P^\infty$ is the image of $x^n$ under the suspension isomorphism. The upshot of this is that after reduction mod $p$ we have that $(k^n-k^j)\equiv 0\mod p$ if and only if $(n-j)\equiv 0\mod p-1$
Finally we form a last set of maps $\theta_j:\Sigma\mathbb{C}P^\infty\rightarrow\Sigma\mathbb{C}P^\infty$, for $j=1,\dots,p-1$ by setting
$$\theta_j=g_1\circ\dots\circ\widehat g_j\circ\dots\circ g_{p-1}$$
where the hat indicates that the map is omitted from the composition. Again you can calculate what each of these maps do on cohomology, but it is easy to see from our previous comments that $\theta_j$ induces multiplication by a mod $p$ unit in degrees congruent to $j$ mod $p-1$ and the $0$-homomorphism in all other degrees.
Therefore we should consider the decomposition
$$H^*\Sigma\mathbb{C}P^\infty\cong \bigoplus_{j=1,\dots,p-1} M_j$$
where $M_j$ contains all the cohomology in degrees congruent to $j$ mod $p-1$. Then $im(\theta_j^*)\subseteq M_j$. The point is that if we now consider cohomology with mod $p$ coefficients, then $im(\theta_j^*)=M_j$, since in the degrees in which $\theta_j^*$ is non-zero it is just multiplication by a mod $p$ unit. What we actually need to do, however, is to consider coefficients in the $p$-local integers $\mathbb{Z}_{(p)}$. Then the map
$$\bigoplus_{j=1,\dots,p-1}\theta_j^* :H^*(\Sigma\mathbb{C}P^\infty;\mathbb{Z}_{(p)})\xrightarrow{\cong} \bigoplus_{j=1,\dots,p-1} M_j$$
realises the previous isomorphism.
The point of all this is that $\theta_j^*$ is induced by an actual map of spaces, and is just not some abstract module decomposition. Therefore, for each $j=1,\dots,p-1$ we let $K_j$ be the mapping telescope
$$K_j=Tel\big(\Sigma\mathbb{C}P^\infty\xrightarrow{\theta_j}\Sigma\mathbb{C}P^\infty\Sigma\mathbb{C}P^\infty\xrightarrow{\theta_j} \dots\big).$$
Then calculating with $\mathbb{Z}_{(p)}$-cohomology we have that $H^*(K_j;\mathbb{Z}_{(p)})\cong M_j$. For each $j$ there is a map $f_j:\Sigma\mathbb{C}P^\infty\rightarrow K_j$ induced by including $\Sigma\mathbb{C}P^\infty$ as the first space in the telescope, and when we form the map
$$f:\Sigma\mathbb{C}P^\infty\xrightarrow{\oplus f_j}\bigvee_{j=1,\dots,p-1}K_j,$$
the sum once again being formed using the suspension coordinate, we find that
$$f^*:\bigoplus_{j=1,\dots,p-1}M_j\cong H^*(\bigvee_{j=1,\dots,p-1}K_j;\mathbb{Z}_{(p)})\xrightarrow{\cong} H^*(\Sigma\mathbb{C}P^\infty;\mathbb{Z}_{(p)})$$
is exactly the isomorphism we wrote down abstractly above.
Now in the integral world this doesn't say much, but after localisation at $p$, we can appeal to the Whitehead theorem, since $f$ is a map between simply-connected spaces which induces an isomorphism on $\mathbb{Z}_{(p)}$-cohomology. Thus $f$ is a p-local homotopy equivalence
$$\Sigma\mathbb{C}P^\infty\simeq \bigvee_{j=1,\dots,p-1}K_j,$$
and we see that localised at the odd prime $p$, the suspension $\Sigma\mathbb{C}P^\infty$ splits as a wedge-sum of $p-1$-pieces, whose cohomology we can easily understand.
Now let us return to the problem at hand. We'll start with $\mathbb{C}P^p$, suspend and include into $\Sigma\mathbb{C}P^\infty$. Then for cellular reasons the homotopy equivalence we've just written down gives us that
$$\Sigma\mathbb{C}P^p\simeq S^3\cup e^{2p+1}\vee S^5\vee S^7\vee\dots\vee S^{2p-1}$$
since these are the only cells of the $K_j$ in these dimensions (to see this just study their cohomological counterparts $M_j$).
Now the first wedge summand is a 2-cell complex formed by attaching a $(2p+1)$-cell to $S^3$ by a map $\alpha:S^{2p}\rightarrow S^3$. Since the Steenrod operations commute with suspension we get
$$\mathcal{P}^1(\sigma x)=\sigma(\mathcal{P}^1x)=\sigma x^p,$$
and it follows from this that $\alpha$ must be non-trivial. Hence there is an element $\alpha\in\pi_{2p}S^3$ which is non-trivial and detected by $\mathcal{P}^1$.