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This is also an incomplete answer but from a different angle than the previous ones. Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$.

EDIT: the answer below only deals with the orientable case $\det(A)=1$. The rational cohomology in the nonorientable case can be handled similarly.

The picture will look qualitatively different depending on whether the eigenvalues have absolute values 1 (corresponds to trace of $A$ equal to 0 or $\pm 2$) or not.

If $tr(A)\ne 0, \pm 2$ then $|\lambda_i(A)|\ne 1$ and the resulting 3-manifold is a solvmanifold. In this case the commutator of $\pi_1(M)$ has rank 2, hence the abelianization of $\pi_1(M)$ has rank 1 and therefore both $H^1(M)$ and $H^2(M)$ have rank 1.
Over $\mathbb Q$ (or $\mathbb R$ ) the generator of $H^1(M)$ comes from the pullback of the generator of the base circle as can be seen from the fact that the original bundle has a section. Using Poincare duality we now easily get the cup product structure on $H^*(M,\mathbb Q)$. There are generators $a\in H^1(M)\cong \mathbb Q, b\in H^2(M)\cong \mathbb Q, c\in H^3(M)\cong \mathbb Q$ and $a^2=0, ab=c$.

The other case when $|\lambda_i(A)=1|$ gives infranilmanifolds. Let me discuss the nil case when $tr(A)=2$ and $\lambda_1=\lambda_2=1$. Then in appropriate coordinates on $T^2$ the mapping matrix $A$ reduces to $\begin{pmatrix} 1&k\\0&1\end{pmatrix}$.

This actually can be viewed as an $S^1$ bundle $S^1\to M\to T^2$ over a 2-torus (made out of the base circle and the first meridian of $T^2$ which is fixed by $A$) . The number $k$ is the Euler class of this circle bundle. You can get cohomology from the Gysin sequence of this circle bundle. Rationally it's very simple. When $k=0$ you just get $T^3$ and when $k\ne 0$ the minimal model of this is given by the free dga $(\Lambda (x,y,z), d)$ where $\deg x=\deg y=\deg z=1$ and $dx=dy=0, dz=xy$. The cohomology of this algebra is the rational cohomology algebra of $M$: $x$ and $y$ are generators of $H^1$, $xz, yz$ are generators of $H^2$, $t=xyz$ is a generator of $H^3$ and the cup products are:

$xy=0, x(xz)=y(yz)=0, x(yz)=-y(xz)=t$.

Here $x$ and $y$ are generators of $H^1(T^2)$ where $T^2$ is the base torus.

The last case is infranil which in appropriate basis corresponds to $A=\begin{pmatrix} -1&k\\0&-1\end{pmatrix}$ or $A=\begin{pmatrix} 0&1\\-1&0\end{pmatrix}$. These are finitely covered by the nilmanifolds (with covers of order 2 or 4 respectively) and need to be handled separately. In this case the commutator again has rank 2 so the answer is similar to the solv case.

This is also an incomplete answer but from a different angle than the previous ones. Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$.

EDIT: the answer below only deals with the orientable case $\det(A)=1$. The rational cohomology in the nonorientable case can be handled similarly.

The picture will look qualitatively different depending on whether the eigenvalues have absolute values 1 (corresponds to trace of $A$ equal to 0 or $\pm 2$) or not.

If $tr(A)\ne 0, \pm 2$ then $|\lambda_i(A)|\ne 1$ and the resulting 3-manifold is a solvmanifold. In this case the commutator of $\pi_1(M)$ has rank 2, hence the abelianization of $\pi_1(M)$ has rank 1 and therefore both $H^1(M)$ and $H^2(M)$ have rank 1.
Over $\mathbb Q$ (or $\mathbb R$ ) the generator of $H^1(M)$ comes from the pullback of the generator of the base circle as can be seen from the fact that the original bundle has a section. Using Poincare duality we now easily get the cup product structure on $H^*(M,\mathbb Q)$. There are generators $a\in H^1(M)\cong \mathbb Q, b\in H^2(M)\cong \mathbb Q, c\in H^3(M)\cong \mathbb Q$ and $a^2=0, ab=c$.

The other case when $|\lambda_i(A)|=1$ gives infranilmanifolds. Let me discuss the nil case when $tr(A)=2$ and $\lambda_1=\lambda_2=1$. Then in appropriate coordinates on $T^2$ the mapping matrix $A$ reduces to $\begin{pmatrix} 1&k\\0&1\end{pmatrix}$.

This actually can be viewed as an $S^1$ bundle $S^1\to M\to T^2$ over a 2-torus (made out of the base circle and the first meridian of $T^2$ which is fixed by $A$) . The number $k$ is the Euler class of this circle bundle. You can get cohomology from the Gysin sequence of this circle bundle. Rationally it's very simple. When $k=0$ you just get $T^3$ and when $k\ne 0$ the minimal model of this is given by the free dga $(\Lambda (x,y,z), d)$ where $\deg x=\deg y=\deg z=1$ and $dx=dy=0, dz=xy$. The cohomology of this algebra is the rational cohomology algebra of $M$: $x$ and $y$ are generators of $H^1$, $xz, yz$ are generators of $H^2$, $t=xyz$ is a generator of $H^3$ and the cup products are:

$xy=0, x(xz)=y(yz)=0, x(yz)=-y(xz)=t$.

Here $x$ and $y$ are generators of $H^1(T^2)$ where $T^2$ is the base torus.

The last case is infranil which in appropriate basis corresponds to $A=\begin{pmatrix} -1&k\\0&-1\end{pmatrix}$ or $A=\begin{pmatrix} 0&1\\-1&0\end{pmatrix}$. These are finitely covered by the nilmanifolds (with covers of order 2 or 4 respectively) and need to be handled separately. In this case the commutator again has rank 2 so the answer is similar to the solv case.

This is also an incomplete answer but from a different angle than the previous ones. Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$.

EDIT: the answer below only deals with the orientable case $\det(A)=1$. The rational cohomology in the nonorientable case can be handled similarly.

The picture will look qualitatively different depending on whether the eigenvalues have absolute values 1 (corresponds to trace of $A$ equal to 0 or $\pm 2$) or not.

If $tr(A)\ne 0, \pm 2$ then $|\lambda_i(A)|\ne 1$ and the resulting 3-manifold is a solvmanifold. In this case the commutator of $\pi_1(M)$ has rank 2, hence the abelianization of $\pi_1(M)$ has rank 1 and therefore both $H^1(M)$ and $H^2(M)$ have rank 1.
Over $\mathbb Q$ (or $\mathbb R$ ) the generator of $H^1(M)$ comes from the pullback of the generator of the base circle as can be seen from the spectral sequences the other answers discuss. Since the pull back commutes with cup product and using Poincare duality this gives the cup product structure on $H^*(M,\mathbb Q)$. You have generators $a\in H^1(M)\cong \mathbb Q, b\in H^2(M)\cong \mathbb Q, c\in H^3(M)\cong \mathbb Q$ and $a^2=0, ab=c$.

The other case gives infranilmanifolds. Let me discuss the nil case when $tr(A)=2$ and $\lambda_1=\lambda_2=1$. Then in appropriate coordinates on $T^2$ the mapping matrix $A$ reduces to $\begin{pmatrix} 1&k\\0&1\end{pmatrix}$.

This actually can be viewed as an $S^1$ bundle $S^1\to M\to T^2$ over a 2-torus (made out of the base circle and the first meridian of $T^2$ which is fixed by $A$) . The number $k$ is the Euler class of this circle bundle. You can get cohomology from the Gysin sequence of this circle bundle. Rationally it's very simple. When $k=0$ you just get $T^3$ and when $k\ne 0$ the minimal model of this is given by the free dga $(\Lambda (x,y,z), d)$ where $\deg x=\deg y=\deg z=1$ and $dx=dy=0, dz=xy$. The cohomology of this algebra is the rational cohomology algebra of $M$: $x$ and $y$ are generators of $H^1$, $xz, yz$ are generators of $H^2$, $t=xyz$ is a generator of $H^3$ and the cup products are:

$xy=0, x(xz)=y(yz)=0, x(yz)=-y(xz)=t$.

Here $x$ and $y$ are generators of $H^1(T^2)$ where $T^2$ is the base torus.

The last case is infranil which in appropriate basis corresponds to $A=\begin{pmatrix} -1&k\\0&-1\end{pmatrix}$ or $A=\begin{pmatrix} 0&1\\-1&0\end{pmatrix}$. These are finitely covered by the nilmanifolds (with covers of order 2 or 4 respectively) and need to be handled separately. In this case the commutator again has rank 2 so the answer is similar to the solv case.

This is also an incomplete answer but from a different angle than the previous ones. Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$.

EDIT: the answer below only deals with the orientable case $\det(A)=1$. The rational cohomology in the nonorientable case can be handled similarly.

The picture will look qualitatively different depending on whether the eigenvalues have absolute values 1 (corresponds to trace of $A$ equal to 0 or $\pm 2$) or not.

If $tr(A)\ne 0, \pm 2$ then $|\lambda_i(A)|\ne 1$ and the resulting 3-manifold is a solvmanifold. In this case the commutator of $\pi_1(M)$ has rank 2, hence the abelianization of $\pi_1(M)$ has rank 1 and therefore both $H^1(M)$ and $H^2(M)$ have rank 1.
Over $\mathbb Q$ (or $\mathbb R$ ) the generator of $H^1(M)$ comes from the pullback of the generator of the base circle as can be seen from the fact that the original bundle has a section. Using Poincare duality we now easily get the cup product structure on $H^*(M,\mathbb Q)$. There are generators $a\in H^1(M)\cong \mathbb Q, b\in H^2(M)\cong \mathbb Q, c\in H^3(M)\cong \mathbb Q$ and $a^2=0, ab=c$.

The other case when $|\lambda_i(A)=1|$ gives infranilmanifolds. Let me discuss the nil case when $tr(A)=2$ and $\lambda_1=\lambda_2=1$. Then in appropriate coordinates on $T^2$ the mapping matrix $A$ reduces to $\begin{pmatrix} 1&k\\0&1\end{pmatrix}$.

This actually can be viewed as an $S^1$ bundle $S^1\to M\to T^2$ over a 2-torus (made out of the base circle and the first meridian of $T^2$ which is fixed by $A$) . The number $k$ is the Euler class of this circle bundle. You can get cohomology from the Gysin sequence of this circle bundle. Rationally it's very simple. When $k=0$ you just get $T^3$ and when $k\ne 0$ the minimal model of this is given by the free dga $(\Lambda (x,y,z), d)$ where $\deg x=\deg y=\deg z=1$ and $dx=dy=0, dz=xy$. The cohomology of this algebra is the rational cohomology algebra of $M$: $x$ and $y$ are generators of $H^1$, $xz, yz$ are generators of $H^2$, $t=xyz$ is a generator of $H^3$ and the cup products are:

$xy=0, x(xz)=y(yz)=0, x(yz)=-y(xz)=t$.

Here $x$ and $y$ are generators of $H^1(T^2)$ where $T^2$ is the base torus.

The last case is infranil which in appropriate basis corresponds to $A=\begin{pmatrix} -1&k\\0&-1\end{pmatrix}$ or $A=\begin{pmatrix} 0&1\\-1&0\end{pmatrix}$. These are finitely covered by the nilmanifolds (with covers of order 2 or 4 respectively) and need to be handled separately. In this case the commutator again has rank 2 so the answer is similar to the solv case.

This is also an incomplete answer but from a different angle than the previous ones. Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$.

The picture will look qualitatively different depending on whether the eigenvalues have absolute values 1 (corresponds to trace of $A$ equal to 0 or $\pm 2$) or not.

If $tr(A)\ne 0, \pm 2$ then $|\lambda_i(A)|\ne 1$ and the resulting 3-manifold is a solvmanifold. In this case the commutator of $\pi_1(M)$ has rank 2, hence the abelianization of $\pi_1(M)$ has rank 1 and therefore both $H^1(M)$ and $H^2(M)$ have rank 1.
Over $\mathbb Q$ (or $\mathbb R$ ) the generator of $H^1(M)$ comes from the pullback of the generator of the base circle as can be seen from the spectral sequences the other answers discuss. Since the pull back commutes with cup product and using Poincare duality this gives the cup product structure on $H^*(M,\mathbb Q)$. You have generators $a\in H^1(M)\cong \mathbb Q, b\in H^2(M)\cong \mathbb Q, c\in H^3(M)\cong \mathbb Q$ and $a^2=0, ab=c$.

The other case gives infranilmanifolds. Let me discuss the nil case when $tr(A)=2$ and $\lambda_1=\lambda_2=1$. There is a 1-eigenvalue and in appropriate coordinates on $T^2$ the mapping matrix $A$ reduces to $\begin{pmatrix} 1&k\\0&1\end{pmatrix}$.

This actually can be viewed as an $S^1$ bundle $S^1\to M\to T^2$ over a 2-torus (made out of the base circe and the first meridian of $T^2$ which is fixed by $A$) . the number $k$ is the Euler class of this circle bundle. You can get cohomology from the Gysin sequence. Rartionally it's very simple. When $k=0$ you just get $T^3$ and when $k\ne 0$ the minimal model of this is given by the dga $(\Lambda (x,y,z), d)$ where $\deg x=\deg y=\deg z=1$ and $dx=dy=0, dz=xy$. The cohomology of this algebra is the rational cohomology algebra of $M$: $x$ and $y$ are generators of $H^1$, $xz, yz$ are generators of $H^2$, $t=xyz$ is a generator of $H^3$ and the cup products are:

$xy=0, x(xz)=y(yz)=0, x(yz)=-y(xz)=t$.

Here $x$ and $y$ are generators of $H^1(T^2)$ where $T^2$ is the base torus.

The last case is infranil which in appropriate basis corresponds to $A=\begin{pmatrix} -1&k\\0&-1\end{pmatrix}$ or $A=\begin{pmatrix} 0&1\\-1&0\end{pmatrix}$. These are finitely covered by the nilmanifolds (with covers of order 2 or 4 respectively) and need to be handled separately. I think in this case the commutator again has rank 2 so the answer is similar to the solv case.

This is also an incomplete answer but from a different angle than the previous ones. Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$.

EDIT: the answer below only deals with the orientable case $\det(A)=1$. The rational cohomology in the nonorientable case can be handled similarly.

The picture will look qualitatively different depending on whether the eigenvalues have absolute values 1 (corresponds to trace of $A$ equal to 0 or $\pm 2$) or not.

If $tr(A)\ne 0, \pm 2$ then $|\lambda_i(A)|\ne 1$ and the resulting 3-manifold is a solvmanifold. In this case the commutator of $\pi_1(M)$ has rank 2, hence the abelianization of $\pi_1(M)$ has rank 1 and therefore both $H^1(M)$ and $H^2(M)$ have rank 1.
Over $\mathbb Q$ (or $\mathbb R$ ) the generator of $H^1(M)$ comes from the pullback of the generator of the base circle as can be seen from the spectral sequences the other answers discuss. Since the pull back commutes with cup product and using Poincare duality this gives the cup product structure on $H^*(M,\mathbb Q)$. You have generators $a\in H^1(M)\cong \mathbb Q, b\in H^2(M)\cong \mathbb Q, c\in H^3(M)\cong \mathbb Q$ and $a^2=0, ab=c$.

The other case gives infranilmanifolds. Let me discuss the nil case when $tr(A)=2$ and $\lambda_1=\lambda_2=1$. Then in appropriate coordinates on $T^2$ the mapping matrix $A$ reduces to $\begin{pmatrix} 1&k\\0&1\end{pmatrix}$.

This actually can be viewed as an $S^1$ bundle $S^1\to M\to T^2$ over a 2-torus (made out of the base circle and the first meridian of $T^2$ which is fixed by $A$) . The number $k$ is the Euler class of this circle bundle. You can get cohomology from the Gysin sequence of this circle bundle. Rationally it's very simple. When $k=0$ you just get $T^3$ and when $k\ne 0$ the minimal model of this is given by the free dga $(\Lambda (x,y,z), d)$ where $\deg x=\deg y=\deg z=1$ and $dx=dy=0, dz=xy$. The cohomology of this algebra is the rational cohomology algebra of $M$: $x$ and $y$ are generators of $H^1$, $xz, yz$ are generators of $H^2$, $t=xyz$ is a generator of $H^3$ and the cup products are:

$xy=0, x(xz)=y(yz)=0, x(yz)=-y(xz)=t$.

Here $x$ and $y$ are generators of $H^1(T^2)$ where $T^2$ is the base torus.

The last case is infranil which in appropriate basis corresponds to $A=\begin{pmatrix} -1&k\\0&-1\end{pmatrix}$ or $A=\begin{pmatrix} 0&1\\-1&0\end{pmatrix}$. These are finitely covered by the nilmanifolds (with covers of order 2 or 4 respectively) and need to be handled separately. In this case the commutator again has rank 2 so the answer is similar to the solv case.