I don't know how explicit we can go, but I'll give it a try. We have to go first through the homotopy-theoretical part.
Since $\{ * \} \subseteq X, \{ * \} \subseteq Y$ are cofibrations, $X \vee Y \subseteq X \times Y$ also is. Let $Z$ be a pointed space and consider the long exact sequence of homotopy for the pair $X \vee Y \subseteq X \times Y$, ie. the sequence
$\ldots \rightarrow [\Sigma ^{2}(X \vee Y), Z] \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X \vee Y), Z] \rightarrow [X \wedge Y, Z] \rightarrow \ldots$,
where $[-,-]$ is the pointed set of homotopy classes of basepoint-preserving maps. Note that for any $n \geq 0$, $\Sigma^{n}(X \vee Y)$ is homeomorphic to $\Sigma^{n}X \vee \Sigma ^{n} Y$. I will not distinguish between the two.
Let $k \geq 1$ and define a map
$\psi ^{k}: \Sigma^{k}(X \times Y) \rightarrow \Sigma^{k}X \vee \Sigma^{k}Y$
$\psi ^{k} = \Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})$,
where $\pi: X \times Y \rightarrow X, Y$ are the projections and $i: X, Y \rightarrow X \vee Y$ are the inclusions. Addition is performed via the suspension structure on $\Sigma^{k}(X \times Y)$, so this is why we require $k \geq 1$. (Observe that even though I denote it by addition this is not necessarily commutative for $k=1$.)
If $j: X \vee Y \hookrightarrow X \times Y$ is the inclusion, then I claim that $\psi ^{k}$ is the left inverse to $\Sigma^{k}j$, ie. $\psi ^{k} \circ \Sigma^{k}j = id_{\Sigma^{k}(X \vee Y)}$. This is important because $\Sigma^{k}j$ are connecting maps in the long exact sequence of homotopy. Indeed, one computes
$\psi ^{k} \circ (\Sigma^{k}j) = (\Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})) \circ \Sigma^{k}j = \Sigma^{k}(i_{X} \pi_{X} j) + \Sigma^{k}(i_{Y} \pi _{Y} j) = \Sigma^{k}(id_{X} \vee const) + \Sigma^{k}(const \vee id_{Y}) \simeq (\Sigma^{k}id_{X} + const) \vee (const + \Sigma^{k}id_{Y}) \simeq \Sigma^{k}id_{X} \vee \Sigma^{k}id_{Y} \simeq id_{\Sigma^{k}X \vee \Sigma^{k}Y}$.
(One can also see this geometrically.) This immediately implies that for all $k \geq 1$ and all $Z$ the $[\Sigma^{k}(X \times Y), Z] \rightarrow [\Sigma^{k}(X \vee Y), Z]$ induced by $j$ is surjective and - by exactness of the long exact sequence - that for all $n \geq 1$ the map $[\Sigma^{n}(X \smash Y), Z] \rightarrow [\Sigma^{n}(X \times Y), Z]$ has zero kernel. In particular, for $k=1$ we have the short exact sequence of groups
$0 \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow 0$
Moreover, the map induced by $\psi^{1}$ splits it and shows that there is a natural isomorphism
$\phi: [\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow [\Sigma(X \times Y), Z]$,
of groups, where the product is only semi-direct, because our groups are not necessarily abelian. This is enough for our purposes, since we also have natural bijections
$[\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y), Z] \times [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), Z]$.
(The second one follows from from the fact that $\vee$ is the direct sum in the category of pointed spaces.) Yoneda lemma establishes that there is an isomorphism
$\theta: \Sigma(X \times Y) \rightarrow _{\simeq} \Sigma(X \smash Y) \vee \Sigma(X) \vee \Sigma(Y) $
in the homotopy category of pointed spaces, ie. a homotopy equivalence that we were after. It takes a little bookkeeping in the above Yoneda-lemma argumentation to see that such map is given by
$\theta = \Sigma(p) + \psi^{i} = \Sigma(p) + \Sigma^{1}(i_{X} \pi_{X}) + \Sigma^{1}(i_{Y} \pi_{Y})$,
where $p: X \times Y \rightarrow X \wedge Y$ is the natural projection. (This is what we get if we start with $id \in [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ and trace it back by all the bijections above to $[\Sigma(X \times Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ - and this is the way to discover the isomorphisms "hidden" by Yoneda lemma.)
I understand that my exposition is far from perfect, but if you would like me to go into more detail over some parts, please comment.
The functor $\Omega$ is the composite of a bunch of functors, which I will now describe. Write $\mathsf{A}$ for the diagram $2\to 3\leftarrow 1$. The inclusion $i_3\colon *\to\mathsf{A}, *\mapsto 3$ induces a left Kan extension $\mathrm{Lan}_{i_3}\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{A},\mathscr{C})$ (informally sending $X\in\mathscr{C}$ to $*\to X\leftarrow*$). Write $\mathsf{Sq}$ for the category $[1]\times[1]$, but instead of $(00)$, $(01)$, $(10)$, and $(11)$, I call the objects $0$, $1$, $2$ and $3$, respectively. Write $i\colon\mathsf{A}\to\mathsf{Sq}$ for the obvious inclusion. This induces a right Kan extension $\mathrm{Ran}_i\colon\mathrm{Fun}(\mathsf{A},\mathscr{C})\to\mathrm{Fun}(\mathsf{Sq},\mathscr{C})$. Finally, the inclusion $i_0\colon *\to\mathsf{Sq}, *\mapsto 0$ induces a restriction functor $i_0^*\colon\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\to\mathscr{C}$. The functor $\Omega$ is the composite
$$
\mathscr{C}\xrightarrow{\mathrm{Lan}_{i_3}}\mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{\mathrm{Ran}_i}\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\xrightarrow{i_0^*}\mathscr{C}.
$$
The functor $i_0^*$ is equivalent to the limit functor $\mathrm{lim}\colon\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\to\mathscr{C}$ (since the limit of a diagram with an initial object equals the value at this initial object), so $i_0^*$ has the constant diagram functor $\Delta\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{Sq},\mathscr{C})$ as left adjoint. The functor $\mathrm{Ran}_i$ is right adjoint to $i^*\colon\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\to\mathrm{Fun}(\mathsf{A},\mathscr{C})$ by definition. Finally, we claim that $\mathrm{Lan}_{i_3}$ is also a right adjoint functor. Let us find a left adjoint.
Write $\mathsf{B}$ for the category $(-2)\leftarrow 2\to 3\leftarrow 1\to (-1)$. There is an inclusion $j\colon\mathsf{A}\to\mathsf{B}, n\mapsto n$. Let $F\colon\mathrm{Fun}(\mathsf{A},\mathscr{C})\to\mathscr{C}$ be the composite
$$
\mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{\mathrm{Ran}_j}\mathrm{Fun}(\mathsf{B},\mathscr{C})\xrightarrow{\mathrm{colim}}\mathscr{C}.
$$
Informally, $F$ sends a diagram $A\to C\leftarrow B$ to $\mathrm{colim}(*\leftarrow A\to C\leftarrow B\to *)$. (This functor is informally the ''main reason'' that the left adjoint to $\Omega$ will be the suspension functor).
We claim that $F$ is left adjoint to $\mathrm{Lan}_{i_3}$. Given a functor $X\colon\mathsf{A}\to\mathscr{C}$, and an object $Y\in\mathscr{C}$, the mapping anima $\mathscr{C}^\mathsf{A}(X,\mathrm{Lan}_{i_3}Y)$ (an ''anima'' is a synonym for ''homotopy type'' that the cool kids sometimes use nowadays --I use it because Qi did, not because I am one of those cool kids myself) can be pictured as the anima of diagrams
where the dashed maps may be freely picked, and the solid arrows are fixed. Let $p_1\colon X_1\to *$, $p_2\colon X_2\to *$, $p_{31}\colon X_3\to*$ and $p_{32}\colon X_3\to *$ be the maps given to us in the diagram $\mathrm{Ran}_j X$, which come with homotopies $p_1\simeq(X_1\to X_3\xrightarrow{P_{31}}*) and $p_2\simeq(X_2\to X_3\xrightarrow{p_{32}}*). The anima of maps $X\to *$ is contractible, as is the anima of maps $*\to Y$. Via a pullback of the anima of diagrams as above along the inclusion of those diagrams where the maps $X_1\to *$, $X_2\to *$, $X_3\rightrightarrows *$ and the homotopies between them are given by $p_1$, $p_2$, $p_{31}$, $p_{32}$ and the homotopies between them, we get a naturally(!) equivalent anima of diagrams
Likewise, we have a natural equivalence from the above anima towards the anima of diagrams
where we no longer require the maps $*\rightrightarrows Y$ to be specified. This anima is naturally equivalent to the anima $\mathscr{C}^\mathsf{B}(\mathrm{Ran}_j X, \Delta Y)$, with $\Delta\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{B},\mathscr{C})$ the constant functor. If you want to be precise about this (for instance, if you worry about naturality), you can look at the anima of natural transformations $H\colon\mathsf{B}\times[1]\to\mathscr{C}$ for which $H(0,-)$ equals $\mathrm{Ran}_j X$ and for which $H(-,1)$ equals a diagram which is equivalent to a diagram equivalent to one in which all objects are $Y$ and all morphisms are an identity morphism on $Y$ (and we require the higher morphisms of this anima to respect these boundary conditions). This anima is equivalent to the anima of functors $\mathsf{B}\star[0]\to\mathscr{C}$ which map $\mathsf{B}$ to $\mathrm{Ran}_j X$ and map the cone point to $Y$ (and for which the higher morphisms respect these boundary condition), via standard arguments. The natural zigzag of maps comparing these two functor anima (induced by a zigzag of maps of diagram shapes, hence why it is natural) witnesses this equivalence. But, the latter subanima of functors $\mathsf{B}\star[0]\to\mathscr{C}$ that we considered is exactly the anima of diagrams we saw above.
So, we have found that this anima is naturally equivalent to $\mathscr{C}^\mathsf{B}(\mathrm{Ran}_j X, \Delta Y)$. This in turn is by the adjunction $\mathrm{colim}\dashv\Delta$ naturally equivalent to the anima $\mathscr{C}(\mathrm{colim}\,\mathrm{Ran}_j X, Y)=\mathscr{C}(FX,Y)$. Therefore, we see that $F$ is indeed left adjoint to $\mathrm{Lan}_{i_3}$.
This means that $\Omega$ is right adjoint to the composition
$$
\mathscr{C}\xrightarrow{\Delta}\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\xrightarrow{i^*}\mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{F}\mathscr{C}
$$
which we claim is the suspension functor. This boils down to showing that, for any $X\in\mathscr{C}$, the suspension $\Sigma X$ is the colimit of the diagram
The colimit of this diagram may be computed in a natural way by successively taking pushouts as indicated below:
But this makes it clear that this colimit is naturally equivalent to $\Sigma X$. Hence $\Sigma\dashv\Omega$.
Now, what about the adjunction unit? The adjunction unit of a composite of adjunctions is an approrpriate composite of adjunction units. It is not hard to compute that the adjunction unit of $F\dashv\mathrm{Lan}_{i_3}$ is given for $X\colon\mathsf{A}\to\mathscr{C}$ by the map
The adjunction unit of $i^*\dashv\mathrm{Ran}_i$ is given for $Y\colon\mathsf{B}\to\mathscr{C}$ by the map
For the adjunction $\mathrm{Lan}_{i_0}\dashv i_0^*$ and $Y\in\mathscr{C}$, the adjunction unit is simply $\mathrm{id}_Y$. So, starting with $X\in\mathscr{C}$, the adjunction unit of $\Sigma\dashv\Omega$ equals the composition
$$
X\xrightarrow{\mathrm{id}} X\xrightarrow{\mathrm{id}\times_{\mathrm{id}}\mathrm{id}} {X\times_X X} \xrightarrow{p_1\times_{p}p_2}{*\times_{\Sigma X}*\simeq\Omega\Sigma X}
$$
where $p\colon X\to \Sigma X$ is the canonical map coming from the pushout square. But this is precisely the map $\eta\colon X\to \Omega\Sigma X$ that you were after.
Best Answer
Quite accidentally I've stumbled on a result that basically answers my question.
Proposition A.16 in the appendix of Hatcher is the following:
Sketch of proof:
The key point is that $X\times Y$ and $X$ are Hausdorff spaces, so compact subsets of these spaces will be compact Hausdorff, and thus normal. We can use this normality to prove the following two lemmas, which combine to give the result.
Let $M(K,U)$ denote a compact-open subbasic set. Then
Then applying the second result to $Q=\Maps(Y,Z)$, with subbasis $V=M(B,U)$, we find that $M(A,M(B,U))$ is a subbasis for $\Maps(X,\Maps(Y,Z))$.
Since $M(A\times B,U)\leftrightarrow M(A,M(B,U))$ under the natural bijection, this proves that the natural bijection is in fact a homeomorphism. $\blacksquare$
This then answers my question, at least to a point where I am satisfied. (I would still be interested in a counterexample to the conclusion of the proposition when the assumption that $X$ is Hausdorff is dropped.)
To get the desired result, we note that when $X$ is Hausdorff, $Y$ is locally compact Hausdorff, the homeomorphism induces a natural homeomorphism $$\Maps_*(X,\Maps_*(Y,Z)) \simeq \Maps_*(X\wedge Y, Z)$$ using that $\Maps$ applied to a quotient map in the contravariant variable or an embedding in the covariant variable gives an embedding.
Then taking $Y=S^1$, we get that the adjunction $$\Maps_*(\Sigma X, Z)\simeq \Maps_*(X,\Omega Z) $$ is a homeomorphism whenever $X$ is Hausdorff.