I can only point to the place where this was originally done (or rather, the latest edition thereof):
Topology and Groupoids by Ronnie Brown
It's a fantastic textbook and easy to read (and cheap, if you buy the electronic copy - the best £5 I've spent). Ideally what you'd do is calculate the equivalent subgroupoid $\Pi_1(S^1,\{a,b,c\})$ where $a,b,c$ are three points in $S^1$, one in each intersection of opens.
I don't think the Seifert-van Kampen theorem follows from these kinds of considerations. Rather, it is the statement that the fundamental groupoid functor $\tau_{\leq 1}$ preserves homotopy colimits. That's because it is left adjoint to the inclusion of 1-groupoids in $\infty$-groupoids (a generalization to any $\infty$-category is in Higher Topos Theory, Prop. 5.5.6.18).
[Removed the part on cohomology because it got me confused!]
Added:
I attempted to explain the long exact sequences in cohomology without using spectra, but that was wrong. Here's a correct explanation. The reduced cohomology of a pointed space $X$ depends only on its stabilization $\Sigma^\infty X$: it is given by $H^n(X;A)=[\Sigma^\infty X,\Sigma^n HA]$ where $HA$ is an infinite delooping of $K(A,0)$. The functor $\Sigma^\infty$ preserves cofiber sequences (being left adjoint). Now if you have a cofiber sequence in a stable category, you get long exact sequences of abelian groups when you apply functors like $[E,-]$ or $[-,E]$.
Added later:
My original answer was correct, but like I said I got confused… Here it is again. Let $A\to B\to C$ be a cofiber sequence of pointed spaces. As you say in your question, you get a fiber sequence of mapping spaces
$Map(C,X)\to Map(B,X)\to Map(A,X)$
for any $X$, because $Map$ transforms homotopy colimits in its first variable into homotopy limits. In its second variable it preserves homotopy limits, so $\Omega Map(A,X)=Map(A,\Omega X)$. Applying to $X=K(G,n)$ gives you the usual long exact sequence in cohomology, but only from $H^0$ to $H^n$.
Best Answer
According to Jahrbuch der Mathematik this theorem is in:
See the (long) review in zbMATH by Erika Pannwitz. The relevant sentence is:
See also footnote 65 on page 19 of: