It seems to be geometrically clear what the "pinching map" $\mu : S^n \to S^n \vee S^n$ looks like, but we have to be very precise.
Let us assume that the basepoint $*$ of $S^n$ lies in the equatorial sphere $S^{n-1}_0 = S^{n-1} \times \{0\} \subset S^n$. We may simply take $* = (1,0,\dots,0)$. Then the quotient $S^n /S^{n-1}_0 $ is certainly homeomorphic to the wedge of two copies of $S^n$, but this does not specify a concrete $\mu$.
Let us first construct a homeomorphism $D^n/S^{n-1} \to S^n$. The construction is technically non-trivial; it is a slight modification of that presented in my answer to Two topology questions regarding quotient $D^n/S^{n-1}$ and homotopy $S^{n-1} \to S^{n} - \{ a,b\}$. We write $S^n = \{ (r,x) \in \mathbb{R} \times \mathbb{R}^n \mid r^2 + \lVert x \rVert^2 = 1 \}$, where $\lVert - \rVert$ denotes the Euclidean norm.
Define
$$p : D^n \to S^n, p(x) =
\begin{cases}
\left(2\lVert x \rVert -1, \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right) & x \ne 0 \\
(-1,0) & x = 0
\end{cases}
$$
This is well-defined because $(2\lVert x \rVert -1) \in [-1,1]$ and
$$ (2\lVert x \rVert-1)^2 + \left\lVert \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right\rVert^2= (2\lVert x \rVert -1)^2 + 1 - (2\lVert x \rVert -1)^2 = 1 .$$
$p$ is continuous because for $x \to 0$ we get $\left\lVert \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right\rVert = \sqrt{1 - (2\lVert x \rVert -1)^2} \to 0$ and $2\lVert x \rVert -1 \to -1$.
For $x \in S^{n-1}$ we have $p(x) = (1,0) = *$.
Next define
$$j : S^n \setminus \{*\} \to D^n \setminus S^{n-1}, j(r,y) =
\begin{cases}
\dfrac{1 + r}{2\sqrt{1- r^2}}y & y \ne (-1,0) \\
0 & y = (-1,0)
\end{cases}
$$
This is well-defined because $-1 < r < 1$ for $(r,y) \in S^n \setminus \{*, (-1,0)\}$ and
$$\left\lVert \dfrac{1 + r}{2\sqrt{1- r^2}}y \right\rVert = \dfrac{1 + r}{2\sqrt{1- r^2}}\left\lVert y \right\rVert = \dfrac{1 + r}{2} < 1 .$$
It is easily verified that $p(j(r,y)) = (r,y)$ for all $(r,y)$ and $j(p(x)) = x$ for all $x \in D^n \setminus S^{n-1}$. Hence $p$ maps $D^n \setminus S^{n-1}$ bijectively onto $S^n \setminus \{*\}$.
$D^n , S^n$ are compact Hausdorff, hence $p$ is a closed map and thus a quotient map. By the above considerations there exists a unique function $h : D^n/S^{n-1} \to S^n$ such that $h \circ \pi = p$, where $\pi : D^n \to D^n/S^{n-1}$ is the standard quotient map. By construction it is a bijection. By the universal property of quotient maps $p$ and $\pi$ both $h$ and $h^{-1}$ are continuous, i.e. $h$ is a homeomorphism.
Geometrically $h$ can be visualized as follows: Each sphere $\Sigma^{n-1}_r = \{ x \in D^n \mid \lVert x \rVert = r \} \subset D^n$ with $0 < r < 1$ is mapped to the sphere $S^{n-1}_r = S^n \cap (\{2r-1\} \times \mathbb R^n) \subset S^n$, $S^{n-1}$ goes to $*$ and $0$ to $(-1,0)$. That is, $D^n$ is slipped like a rubber glove over $S^n$ beginning from the left (first coordinate $-1$) to the right (first coordinate $1$).
The equatorial ball $D^{n-1}_0 = D^{n-1} \times \{0\} \subset D^n$ is mapped by $h$ onto the equatorial sphere $S^{n-1}_0$. The other level balls $D^{n-1}_t = D^n \cap (\mathbb R^{n-1} \times \{r\})$ are mapped to "inclined subspheres" of $S^n$ which all meet in $*$.
Let us next observe that in definition 1 we can replace $I$ by $J = [-1,1]$ and obviously obtain an isomorphic group, i.e. we may take
$$\pi_n(X,x_0) = [(J^n,\partial J^n),(X,x_0)]$$
with addition defined by $ [f] +[g)] = [f+g]$, where
$$f+g: J^n \to X, (f+g)(x_1,\ldots,x_n) = \begin{cases}f(2x_1+1,x_2,\ldots,x_n) & x_1 \le 0 \\ g(2x_1-1,x_2,\ldots,x_n) & x_1 \ge 0 \end{cases}$$
There is an obvious homeomorphism $\phi : J^n \to D^n$; see my answer to $(D^n\times I,D^n \times 0)$ and $(D^n \times I, D^n \times 0 \cup \partial D^n \times I)$ are homeomorphic (note that $J^n$ and $D^n$ are the closed unit balls with respect to the maximum-norm $\lVert - \rVert_\infty$ and the Euclidean norm). It maps $J^{n-1}_0 = J^{n-1}\times \{0\}$ to $D^{n-1}_0$.
Consider the quotient map $q = p \circ \phi : J^n \to S^n$. Split $J^n$ into the two cuboids $J^n_\pm = \{(x_1,\ldots, x_n) \mid \pm x_1 \ge 0\}$. Then $q$ induces quotient maps $q_\pm : J^n_\pm \to S^n, q_\pm(x_1,\ldots, x_n) = q(2x_1 \mp 1, x_2,\ldots, x_n)$. Since they agree on $J^{n-1}_0$, they can be pasted to a map $Q : J^n \to S^n \vee S^n$. By construction $Q(\partial J^n) = *$, thus $Q$ induces $\mu : S^n \to S^n \vee S^n$. This is the desired explicit pinching map. By construction $\mu(S^{n-1}_0) = \{*\}$.
It is now a routine exercise to verify that your bijection $\mathcal J$ is a group homomorphism.
Best Answer
The answer is no, not even $\pi_1$ does:
If $X$ is a metric space and $Y$ any topological space, then a map $f\colon X\to Y$ is continuous if and only if for every convergent sequence $(x_n)_{n\in\mathbb{N}}$ in $X$, the sequence $(f(x_n))_{n\in\mathbb{N}}$ converges with limit point $f(\lim_n x_n)$. Furthermore, if $X$ is compact, the closure of the set underlying a convergent sequence is closed and countable. Thus, such an $X$ is naturally isomorphic to the filtered colimit of all of its countable, closed subsets. Fixing a point $x_0\in X$, we may equivalently restrict the indexing category to the set of countable, closed subsets containing $x_0$.
Now take a space as easy as $X = S^1$ with any point $x_0\in X$. We have $\pi_1(A,x_0) = 1$ for each closed and countable subset $A\subset X$, but $\pi_1(X,x_0) = \mathbb{Z}$, of course.