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.