For $p\neq 2$ the answer is yes. It's an easy computation usinghomology decompositions in the sense of Eckmann–Hilton the well known exact sequence
$$\operatorname{Ext}(A,\pi_{n+1}(X))\hookrightarrow [M(A,n),X]\twoheadrightarrow \operatorname{Hom}(A,\pi_n(X)),$$
and the computation
$$\pi_{n+1}(M(A,n))=A\otimes\mathbb{Z}/2.$$
Here $M(A,n)$ is the Moore spectrum whose homology is the abelian group $A$ concentrated in degree $n$.
As you point out, it is easy to check that
$$H_n(M(A,s)\wedge M(B,t))=
\begin{cases}
A\otimes B,&n=s+t,\\
\operatorname{Tor}_1(A,B),&n=s+t+1,\\
0,&\text{otherwise}.
\end{cases}$$
Therefore, $M(A,s)\wedge M(B,t)$ can be obtained as the homotopy cofiber of a map
$$f\colon M(\operatorname{Tor}_1(A,B),s+t)\longrightarrow M(A\otimes B,s+t)$$
which is trivial in homology $H _{*}(f)=0$.
Suppose for instance that $A$ and $B$ are finite and do not have $2$-torsion. Then the previous short exact sequence shows that homology induces an isomorphism
$$H _{s+t}\colon [M(\operatorname{Tor}_1(A,B),s+t), M(A\otimes B,s+t)]\cong \operatorname{Hom}(\operatorname{Tor}_1(A,B),A\otimes B).$$
Therefore $f$ must be null-homotopic, so
$$M(A,s)\wedge M(B,t) \simeq
M(A\otimes B,s+t)\vee M(\operatorname{Tor}_1(A,B),s+t+1).$$
If you take $A=\mathbb{Z}/p^i$ and either $B=\mathbb{Z}/p^j$ or $B=\mathbb{Z}/p^i$ you always obtain the same thing on the right.
For $p= 2$ the answer is no. The spectrum $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ is the mapping cone of
$$4\colon M(\mathbb{Z}/2,0)\longrightarrow M(\mathbb{Z}/2,0)$$
which is knonw to be null-homotopic, actually $[M(\mathbb{Z}/2,0),M(\mathbb{Z}/2,0)]\cong\mathbb{Z}/4$ generated by the identity. Hence
$$M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)\simeq M(\mathbb{Z}/2,0)\vee M(\mathbb{Z}/2,1).$$
In particular the action of the Steenrod algebra on the mod 2 homology of $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ is trivial.
On the other hand, it is known that the Steenrod algebra mode 2 acts on the mod 2 homology of $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/2,0)$ in a non-trivial way, i.e. the first Steenrod operation sends the 0-dimensional generator to the 1-dimensional generator. Therefore $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ and $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/2,0)$ cannot be homotopy equivalent.
As for references, see Hatcher's book. This book doesn't deal with spectra but since we are working with finite-dimensional CW-complexes you can assume that we are in the stable range.
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
Let's extract a clear question (about spectra in general) from your question, and then answer it. Let $E$ be any spectrum.
There is the degree $p$ map $p:S\to S$ from the sphere spectrum to itself. Smashing with $E$, this gives a map $E\to E$; we also call this map $p$. It multiplies elements of $\pi_n(E)$ by $p$. Denote its spectrum cofiber by $E/pE$. The cofibration sequence $E\to E\to E/pE$ is also a fibration sequence, because we are talking about spectra. From the exact sequence $$ \dots \to \pi_n(E)\to \pi_n(E)\to \pi_n(E/pE)\to \pi_{n-1}(E)\to \pi_{n-1}(E)\to \dots $$ you get what I think you want: an exact sequence $$ 0\to coker(p)\to \pi_n(E/pE)\to ker(p)\to 0 $$ where the group on the left is $$ coker (p:\pi_n(E)\to \pi_n(E))=\pi_n\otimes \mathbb F_p $$ (it could also be called $\pi_n(E)/p$) and the group on the right is $$ ker (p:\pi_{n-1}(E)\to \pi_{n-1}(E))=Tor(\pi_{n-1},\mathbb F_p) $$ (I think you are also calling it $\pi_{n-1}(E)[p]$.)
$\pi_n(E/pE)$ is called the $n$th mod $p$ homotopy group of $E$.
Curiously, it can have elements of order $4$ if $p=2$.