Ok, let me try to give you a proof of something that is a lot stronger than what you asked for, but which hopefully is a bit more natural. I am basically going to smother the problem under the abstract nonsense, so I recommend you look at special cases of this proof, e.g. when $X=S^n$ in pointed spaces, in order to get intuition.

Recall that an associative monoid in an ∞-category $D$ with all finite products is a functor
$$M:\Delta^{op}\to D$$
such that for all $[n]\in \Delta^{op}$ the map
$$M([n])\xrightarrow{\prod M(e_i)} \prod_{1\le i\le n} M([1])$$
is an equivalence, where $e_i:[1]\to [n]$ is the map picking out the edge $\{i-1,i\}$ (this is known as the *Segal condition*). Hence, an associative comonoid in an ∞-category with all finite coproducts $C$ is a functor
$$M:\Delta\to C$$
such that
$$\coprod_{1\le i\le n}M([1])\xrightarrow{\coprod M(e_i)} M([n])$$
is an equivalence. Moreover, by postcomposing with the map $C\to hC$, we see that an associative comonoid in $C$ induces an associative comonoid in $hC$.

What I will prove is that, if $C$ is a pointed ∞-category with finite colimits, $\Sigma X$ is an associative comonoid for all $X$ (the same proof, maybe more recognizably, will prove that $\Omega X$ is a group object for all $X$).

Let us consider the category $\Delta_+$ of all finite (possibly empty!) totally ordered sets. This contains two full subcategories we are going to use: $\Delta$ (the category of finite nonempty totally ordered sets) and $\Delta^1$ (the full subcategory spanned by $[0]$ and by the empty set $\varnothing$).

We can construct a functor $M_0:\Delta^1\to C$ sending $[0]$ to $0$ and $\varnothing$ to $X$. Let $M_+$ be the left Kan extension of $M_0$ to $\Delta_+$.

**Lemma** The restriction of $M_+$ to $\Delta$ is an associative comonoid $M$ such that $M([1])\cong \Sigma X$.

*Proof.* Unwrapping the standard formula for left Kan extensions we obtain
$$ M([n])= \mathrm{colim}_{i\in \Delta^1_{[n]/}} M_0(1) = 0\amalg_X 0\amalg_X \cdots \amalg_X 0$$
where in the formula there are $n+1$ 0s. In particular $M([1])=\Sigma X$. For the sake of clarity of notation let me prove the Segal condition only for $n=2$. Then we need to prove the map
$$(0\amalg_X 0)\amalg (0\amalg_X 0) \to 0\amalg_X 0 \amalg_X 0$$
induced by the inclusions of the corresponding summands is an equivalence. But this is simply the associativity of the pushout. $\square$

Finally let me say a few words about how to prove $\Sigma X$ has the structure of a cogroup and not simply a comonoid. If we let
$$m:M([1])\to M([2])\cong M([1])\amalg M(1)$$
be the map induced by the edge $\{0,2\}$ (the "comultiplication"), the condition of being a cogroup is equivalent as asking the map
$$M([1])\amalg M([1])\xrightarrow{m\amalg i_1} M([1])\amalg M([1])$$
is an equivalence. To check this is the case for the $M$ I constructed above is left as an exercise for the reader.