Unfortunately I don't see an "obvious" structure map on the wedge that you describe. It looks more like the smash product for symmetric spectra and doesn't work without the symmetric group action. You'd need to describe maps
$$
S^1 \wedge X_p \wedge Y_q \to \bigvee_{r+s=p+q+1} X_r \wedge Y_s
$$
and it's not clear whether one should use the structure map of $X$ or $Y$ (possibly first applying the twist), or perhaps you had an application of the pinch-multiplication in mind?
So far as the two definitions of smashing with $S^1$, you need to be careful; I'm not sure that both of these definitions are functorial.
In case you hadn't already investigated: for questions about constructing a smash product on these types of spectra, Adams' "Stable homotopy and generalized homology" describes in detail a construction of "handicrafted" smash products by starting with $X_0 \wedge Y_0$ and making a sequence of choices about whether to apply the structure map to the left- or right-hand factor. The proof that it is independent of these choices and gives rise to a symmetric monoidal structure (all after passing to the homotopy category) has been superseded technologically and I wouldn't recommend reading it in detail.
Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).
The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are:
$$ (x_1,y,0)\sim (x_2,y,0);$$
$$ (x,y_1,1)\sim (x,y_2,1).$$
The reduced version also collapses $x_0\ast Y$ and $X\ast y_0$.
So the additional relations are:
$$ (x_0,y,t)\sim (x_0,y_0,0);$$
$$ (x,y_0,t)\sim (x_0,y_0,0).$$
We can derive further relations too:
$$(x,y,0)\sim(x_0,y,0)\sim(x_0,y_0,0);$$
$$(x,y,1)\sim (x,y_0,1)\sim(x_0,y_0,0).$$
The smash product is gotten from $X\times Y$ by collapsing $X\times y_0$ and $x_0\times Y$. The suspension of that can be thought of as $X\times Y\times I$, with the relations:
$$ (x,y_0,t)\sim (x_0,y_0,t);$$
$$ (x_0,y,t)\sim (x_0,y_0,t);$$
$$ (x,y,1)\sim (x_0,y_0,1);$$
$$ (x,y,0)\sim (x_0,y_0,0).$$
The reduced suspension adds the relation
$$ (x_0,y_0,t)\sim (x_0,y_0,0).$$
Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,
$$ (x,y_0,t)\sim (x_0,y_0,0);$$
$$ (x_0,y,t)\sim (x_0,y_0,0);$$
$$ (x,y,0)\sim (x_0,y_0,0);$$
$$ (x,y,1)\sim (x_0,y_0,0).$$
Best Answer
The best way to understand the smash product is by its universal properties. One comes from its expression as a quotient. A map $X \times Y \rightarrow Z$ factors through $X \wedge Y$, if and only if, $X \vee Y \subset X \times Y$ is mapped to a single point. This is a useful criterion to create maps out of smash products.
Another useful universal property smash products have is that they satisfy something like a tensor-hom adjunction in the category of pointed spaces. We have $\operatorname{Map_*}(X \wedge Y , Z) \cong \operatorname{Map}(X, \operatorname{Map}(Y,Z))$. This follows from the usual adjunction between product and hom in the unpointed category plus the universal property in the pragraph above, or explicitly $((x,y) \rightarrow f(x,y)) \rightarrow (x \rightarrow (y \rightarrow f(x,y)))$. So from this perspective, the smash product is just the thing that is adjoint to pointed mapping spaces. This is a very important perspective, for example it leads us to studying loop spaces, because maps from a suspension to $Z$ are the same as maps from the original space to $\Omega Z$. From there, one is very close to discovering Puppe sequences, one of the most important results in elementary algebraic topology.
For the record, some people mistakenly say that smash product is the categorical product in the category of pointed spaces. This is wrong. In fact, the categorical product is still the normal product of spaces. Perhaps what causes this confusion for people is that in the category of sets, the categorical product is also the adjoint to hom, but this is not true in general as we've shown.