The most elementary way I know of is to construct its universal cover. First, some preliminaries. We consider any preordered set as a topological space where the open sets are the downward closed sets; a map between two preordered sets is then continuous iff it is order-preserving. The following Lemma is very convenient for thinking about homotopies of such maps:
Lemma: Let $P$ and $Q$ be preordered sets and $f,g:P\to Q$ be order-preserving maps such that $f(x)\leq g(x)$ for all $x\in P$. Then $f$ and $g$ are homotopic.
Proof: Define $H:P\times [0,1]\to Q$ by $H(x,t)=f(x)$ for $t\in [0,1)$ and $H(x,1)=g(x)$. If $U\subseteq Q$ is downward closed, then $$H^{-1}(U)=f^{-1}(U)\times[0,1)\cup g^{-1}(U)\times\{1\}=f^{-1}(U)\times[0,1)\cup g^{-1}(U)\times[0,1]$$ since $g^{-1}(U)\subseteq f^{-1}(U)$. Since $f^{-1}(U)\times[0,1)\cup g^{-1}(U)\times[0,1]$ is clearly open in $P\times[0,1]$, this shows $H$ is continuous.
Corollary: Let $P$ be a preordered set with an element $p$ which is comparable to exactly one other element $q$ of $P$. Then $P$ deformation-retracts onto $P\setminus\{p\}$.
Proof: Define $f:P\to P$ by $f(p)=q$ and $f(x)=x$ for $x\neq p$. Then $f$ is order-preserving and either $f\leq 1_P$ or $f\geq 1_P$ depending on whether $q\leq p$ or $q\geq p$. Either way, by the Lemma, $f$ is homotopic to $1_P$ and this gives a deformation retraction from $P$ to $P\setminus\{p\}$.
Let's now turn to the pseudocircle $S$. We write $S=\{a,b,c,d\}$ with $a,b\leq c,d$.
The universal cover can then be described as $S\times\mathbb{Z}$, with the order given by $(a,n)\leq (d,n-1),(c,n)$ and $(b,n)\leq (c,n),(d,n)$. (Draw a picture of this poset! It should look like a zigzag path with $a$'s and $b$'s alternating on bottom and $c$'s and $d$'s alternating on top.) The covering map $p:S\times\mathbb{Z}\to S$ is then just the projection; it is easy to verify that it is indeed a covering map.
I now claim that $S\times\mathbb{Z}$ is weakly contractible, so that the higher homotopy groups of $S$ are trivial. Every compact subset of $S\times\mathbb{Z}$ is finite, so it suffices to show that $S$ is covered by an increasing sequence of contractible finite subsets. To prove this, note first that $\{(a,0),(c,0)\}$ deformation-retracts to $\{(a,0)\}$ by the Corollary since $(a,0)\leq (c,0)$. Similarly $\{(a,0),(c,0),(b,0)\}$ deformation-retracts to $\{(a,0)(c,0)\}$ since $(b,0)\leq (c,0)$ and $(b,0)$ is not comparable to $(a,0)$. In the same way $\{(a,0),(c,0),(b,0),(d,0)\}$ deformation-retracts to $\{(a,0),(c,0),(b,0)\}$ since $(d,0)$ is only comparable to $(b,0)$. We can continue adding points one-by-one in this way: $(a,1)$ is only comparable to $(d,0)$, then $(c,1)$ is only comparable to $(a,1)$, and so on. We can also add points "in the other direction": $(d,-1)$ is only comparable to $(a,0)$, and then $(b,-1)$ is only comparable to $(d,-1)$, and so on. So, we can add on all of the points of $S\times\mathbb{Z}$ one by one, maintaining a contractible space at each step.
(In fact, with a bit more work, you can show it's possible to glue together all these deformations in sequence continuously to show that $S\times\mathbb{Z}$ is contractible, not just weakly contractible.)
Alternatively, using a bit of heavier machinery from homotopy theory, you can show quickly and directly that there is a weak homotopy equivalence $f:S^1\to S$. Specifically, pick two points $p,q\in S^1$ and let $f$ map $p$ to $c$, $q$ to $d$, and the two open arcs between $p$ and $q$ to $a$ and $b$ respectively. Note that $S$ is covered by the open sets $U=\{a,b,c\}$ and $V=\{a,b,d\}$ which are both contractible (they deformation-retract onto $\{c\}$ and $\{d\}$ respectively, either using the Corollary twice or directly using the Lemma). Also, $f^{-1}(U)=S^1\setminus\{q\}$ and $f^{-1}(V)=S^1\setminus\{p\}$ are contractible, so $f$ restricts to homotopy equivalences $f^{-1}(U)\to U$ and $f^{-1}(V)\to V$. Moreover, $f$ also restricts to a homotopy equivalence $f^{-1}(U\cap V)\to U\cap V$, since $U\cap V=\{a,b\}$ is discrete and $U\cap V$ is a disjoint union of two arcs, one mapping to each of $a$ and $b$. It follows by the theorem below that $f:S^1\to S$ is a weak homotopy equivalence.
Theorem: Let $f:X\to Y$ be a continuous map and let $C$ be an open cover of $Y$ such that for each nonempty finite $F\subseteq C$, $f$ restricts to a weak homotopy equivalence $f^{-1}(U_F)\to U_F$, where $U_F$ is the intersection of all the elements of $F$. Then $f$ is a weak homotopy equivalence.
You can find a proof of (a slight strengthening of) this theorem as Theorem 6.7.11 in tom Dieck's Algebraic Topology, for instance.
Best Answer
The observations allow you to argue much in the same way you do for $S^1$. For the case of the fundamental group, you can construct a "universal cover" $X$ for $\mathbb{S}$ called the digital line. The space $X$ can be viewed as the set of integers $\mathbb{Z}$ with topology generated by the sets $\{2n\}$ and $\{2n,2n+1,2n+2\}$. The "covering map" $p \colon X \to \mathbb{S}$ sends even numbers to either $a$ or $b$ and odd numbers to either $c$ or $d$. (You should be able to figure out the specifics here. We want $X$ to wrap around $\mathbb{S}$, much like we have $\mathbb{R}$ wrapping around $S^1$.) The usual argument for computing the fundamental group of $S^1$ applied in this setting.
The case of higher homotopy groups is similar.
Once you have that, then it suffices to check that your map $f$ above sends a generator of $\pi_1(S^1)$ to a generator of $\pi_1(\mathbb{S})$.