In multilinear algebra a bilinear map $M\times N\to P$, where $M,N,P$ are modules over a ring $R$, is represented by a linear map $M\otimes_R N\to P$: we need the tensor product to talk about bilinearity without leaving the category of modules and linear maps. When $A=M=N=P$, this fits $A$ with a product structure $\mu$, i.e. the structure of an $R$-algebra.
In the case of a cohomology theory $E^\cdot$ with values in a category of $R$-modules, we know that the cohomology module of a space comes with a natural grading. If we write $F\otimes G$ for the functor $X\mapsto F(X)\otimes G(X)$, where $F,G$ takes values in a category with tensor products, a product structure* on a cohomology theory is really a natural transformation $E^\cdot\otimes E^\cdot\to E^\cdot$ of $R$-linear maps of graded modules, which decomposes into morphisms $E^k\otimes E^l\to E^{k+l}$.
We want these morphisms to be represented on the level of spectra, and as in the case of multilinear maps we need the smash product $\wedge$ to talk about "bilinear maps" $E^\cdot\wedge F^\cdot\to G^\cdot$ without leaving the category of spectra. The definition itself is straightforward: the smash product functor is left-adjoint to $\hom$.
The proof of its existence, however, requires a construction, which turns out to be really difficult to provide: there are now many of them, and as far as I know they all have both strengths and weaknesses. Some, for instance, are only well behaved in "the" homotopy category of spectra, not on the actual level of spectra. So to answer the question on the last line to the best of my knowledge: yes, this should at the very least be what we want.
*For $E^\cdot$ ordinary singular cohomology with values in an $R$-algebra $A$, the product structure on $E^\cdot$ is induced by the coproduct on chain complexes $C_\cdot=C_\cdot(X,Y;A)$, the tensor product of cochains $\xi,\eta$, and the $R$-algebra structure $\mu$ on $A$, i.e. the composition $C_\cdot\to C_\cdot\otimes_R C_\cdot\stackrel{\xi\otimes\eta}{\to} A\otimes_R A \stackrel{\mu}{\to} A.$