Let $X$ be your space. Another possibility, not using van Kampen theorem (explicitely), is to glue a 3-cell inside each sphere. Because the fundamental group depends only on the 2-skeleton, $\pi_1(X)$ is isomorphic to the fundamental group of our new space $\tilde{X}$. Now, $\tilde{X}$ clearly retracts by deformation on a circle, hence $$\pi_1(X) \simeq \pi_1(\tilde{X}) \simeq \pi_1(\mathbb{S}^1) \simeq \mathbb{Z}.$$
Let $$ X= (\mathbb S^2\cup[0, 1])/\{N ∼ 0, S ∼1\},\qquad \qquad Y=\mathbb S^2/\{N ∼ S\}.$$
Then $f\colon X\to Y$ is easy to define. Following your intuition of retracting the interval to a point, just set $f|_{\mathbb S^2}=1_{\mathbb S^2}$ and $f(t)=N=S$ for all $t\in [0,1]$.
Now we need to define $g\colon Y\to X$. The advantage of regarding $\mathbb S^2$ as a union of a pair of (unit) disks, is that on each disk we can parameterize using polar co-ordinates, with $S$ or $N$ as the origin.
The idea now is simple: Firstly $g\colon N,S\mapsto 0.5\in [0,1]$. Then for a point $(r,\theta)$, on either disk of $Y$, if $r\leq0.5$ we map $$g\colon (r,\theta)\mapsto 0.5\pm r\in [0,1].$$
Here the sign $\pm$ is determined by which disk $(r,\theta)$ lies on.
Finally if $r\geq0.5$ we map $$g\colon (r,\theta)\mapsto (2r-1,\theta).$$
Now you have to construct the homotopies $g\circ f ∼ id_X$ and $f \circ g ∼ id_Y$. Let me know if you need help with that.
Let's start with the composition $f \circ g$. This collapses points of radius less than $0.5$ to $N=S$, and applies the function $2r-1$ to radii greater than $0.5$.
We need to construct $H_t\colon Y\to Y$ for $t\in [0,1]$ so that $H_0=1_Y$ and $H_1=f \circ g$.
It makes sense then for $H_t$ to collapse points of radius less than $t/2$ to $N=S$. Then radii from $t/2$ to $1$ should get mapped to radii from $0$ to $1$, which we can do linearly:
$$H_t(r,\theta)=H_t\left(\left(\frac {r-\frac t2}{1-\frac t2},\theta\right)\right),\qquad r\geq t/2,$$
and $H_t(r,\theta)=N=S$ if $r\leq t/2$.
Now check these two definitions agree at the boundary $r=t/2$, and that $H_0=1_Y$ and $H_1=f \circ g$. Check also that points of radius $1$ get mapped to points of radius $1$ for all $t$, as this is needed to ensure continuity where the disks are glued.
Best Answer
further explanation from the comments above:
We want to use the corollary you stated in the question. Pick $U = S_1^2 \sqcup_{p_{-}\mapsto 0} [0,1)$ and $V= S_2^2\sqcup_{p_+\mapsto 1} (0,1]$ where you glue the poles in the same way you described. These are open sets with respect to the quotient topology since their preimage is the union of one sphere with the half-open interval, which is open with respect to the subspace topology of $[0,1]$.
Now the intersection $U\cap V= (0,1)$ is path connected and nonempty. If we can show that $U$ and $V$ have trivial fundamental group, we are done. The symmetry of the situation (more specifically the fact that $U$ and $V$ are homemorphic) allows us to prove it for just one of the subspaces $U$ or $V$.
We know that if two spaces are homotopy equivalent, they have the same fundamental group. The idea is to now contract the line $[0,1)$ glued to the sphere to the single point $\{0\} \sim p_{-}$ which is the south pole on $S_1^2$. This is even stronger than a homotopy equivalence, we actually show that the $S_1^2$ is a deformation retract of $U$.
For this, we just define a map $H\colon U\times [0,1] \to U$ via $$H(x,t) = \begin{cases} x, & x \in S_1^2 \\ (1-t)x, & x \in [0,1]\end{cases}.$$
Note that $H$ is continuous and the identity map on $S_1^2$ for all $t \in [0,1]$, it leaves this subspace invariant for all $t$. Furthermore, $H(.,0)$ is the identity map on $U$ and $H(.,1)$ maps all points of $U$ to $S_1^2$. This shows that $S_1^2$ and $U$ are homotopy equivalent (even a deformation retract). Since $S_1^2$ has trivial fundamental group, $U$ and $V$ do aswell and your corollary is applicable.