Here are some more partial observations, overlapping some of the others. The result won't be completely explicit, as there are too many case distinctions for a clean answer, but hopefully at least the outline is right.

The additive structure can be obtained from the Mayer–Vietoris sequence associated to the covering of your mapping torus $Y$ by two cylinders $U$ and $V$ homeomorphic to $T \times (0,1)$, which on rearrangement will yield (the cohomological analogue of) the result quoted from Hatcher.

The map $H^1(U) \oplus H^1(V) \to H^1(U \cap V)$ is given by $f\colon (a,b) \mapsto (\varphi(b)-a,b-a)$, if we identify $H^1$ of all the solid tori in sight with $\mathbb Z^2$. The contribution of this kernel to $H^1(Y)$ is given by pairs $(a,b)$ with $a = b = \varphi(b)$; in any event, this summand is free abelian. There is another $\mathbb Z$ summand arising as the cokernel of $\delta\colon H^0(U \cap V) \to H^1(Y)$.

The map $H^2(U) \oplus H^2(V) \to H^2(U \cap V)$ is similarly given by $g\colon (m,n) \mapsto (\det(\varphi)n-m,n-m)$ if we identify $H^2$ of each solid torus with $\mathbb Z$. Thus $H^3(Y) = \mathrm{im}\, \delta$ is $\mathbb Z$ if $\det \varphi = 1$ and $\mathbb Z / 2\mathbb Z$ if $\det \varphi = -1$.

Now $H^2(Y)$ fits into a short exact sequence

$$0 \to \mathrm{coker}\, f \to H^2(Y) \to \ker g \to 0.$$

As $\ker g$ is free abelian (either $\mathbb Z$ or $0$ depending on the sign of $\det g$, this sequence will split, with all torsion coming from $\mathrm{coker}\, f$.

The product $H^2(Y) \times H^1(Y) \to H^3(Y)$ will be determined on the free summands by Poincar'e duality if $\det \varphi = 1$. The product of any element with an element $\delta a$ of $\mathrm{im}\,\delta \leq H^1(Y)$ will also be determined by $b \smile \delta a = \delta (b \cdot a)$, where the action of $H^*(Y)$ on $H^*(U \cap V)$ is by restriction followed by cup product (I may be getting the left- vs. right-sidedness wrong—otherwise one picks up a sign—but the Mayer--Vietoris connecting map is, in general, $H^*(Y)$-linear, for $Y$ the covered space.). The same reasoning also covers products with an element of the image of $g$.

I think these considerations together determine most of the product structure, although they do not say anything about $H^1 \times H^2$ on the free parts arising from $\ker f$ and $\ker g$, if $Y$ is nonorientable, nor much about $\ker f \times \ker f$. It might be worth working out a specific example or two to see how bad the remaining ambiguities are.