Your $H_3$ calculation is correct because $H_2(A\cap B)=0$. The $H_2$ calculation is a little fishy. Here is the relevant fragment of the MV sequence:
$$0\to H_2(X)\to H_1(A\cap B) \to H_1(A)\oplus H_1(B),$$
which turns into
$$0\to H_2(X) \to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2.$$
There is no reason to conclude that $H_2(X)=\mathbb Z_2$. Indeed, based on the topology, the map $\mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2$ should be given by $x\mapsto (x,x)$, which is injective. Hence $H_2(X)=0$.
But now you have enough to figure out $H_1(X)$. This is easiest if you use reduced homology. Then, similarly to what you've written you get
$$0\to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2\to H_1(X)\to 0.$$
Thus $H_1(X)\cong (\mathbb Z_2\oplus\mathbb Z_2)/\langle (x,x)\rangle\cong \mathbb Z_2$.
An alternative approach is to use cellular homology. This is basically $\mathbb{RP}^2$ with two $3$-cells attached. The cellular chain complex is
$$0\to \mathbb Z^2\overset{0}{\to} \mathbb Z\overset{\times 2}{\to}\mathbb Z\overset{0}{\to}\mathbb Z\to 0.$$
You can deduce these maps since you know what they are for $\mathbb{RP}^3$, and this is essentially $\mathbb{RP}^3$ with an additional $3$ cell attached in the same way. Taking the homology gives the same answer as Mayer-Vietoris.
Although you've got the answer by yourself, I would like to write an answer solving the problem with cellular homology, so that someone who asks the same question can find an answer here. I solved this problem a few month ago in an Algebraic Topology course as an exercise.
Proof:
Let $X = S^1\times S^1/ \sim$ be the space with the identifications:
$$(e^{2\pi i/m}z,x_0)\sim (z,x_0)$$
$$(x_0,e^{2\pi i/n}z)\sim (x_0,z)$$
Like @Berci said, you should imagine this space as a grid of $m$ and $n$ lines, i.e. there are $m$ vertical and $n$ horizontal repititions:
(OK. The picture is not the nicest one, but it's enough to induce an imagination.)
$X$ consists of one 0-cell ($x_0$ is $e_1^0$), two 1-cells ($a$ is $e_1^1$, $b$ is $e_2^1$) and one 2-cell (we all it $e_1^2$).
The attaching map identifies $x\in \partial D_1^2$ with $a^nb^ma^{-n}b^{-m}$.
This implies the cellular chain complex
$$0\to \mathbb{Z}[e_1^2]\overset{\partial_2=0}{\longrightarrow} \mathbb{Z}[a] \oplus\mathbb{Z}[b]\overset{\partial_1=0}{\longrightarrow} \mathbb{Z}[x_0]\to 0.$$
This implies
$$H_p(X) =
\begin{cases}
\mathbb{Z}\mbox{ for } p=0,2 \\
\mathbb{Z}^2\mbox{ for } p=1 \\
0\mbox{ for } p>2
\end{cases}.$$
Otherwise you can just see, that the space $X$ is still a Torus (cf. remark above). So it is not surprising, that we've got the homology group of the Torus.
Best Answer
Method 1: Group action
Note that $X=(S^2\times S^1)/\Bbb Z_2$, where the action of $\Bbb Z_2$ is generated by $(x,z)\mapsto (-x,-z)$. We can also rewrite $X$ as a quotient space of $S^2\times\Bbb R$ using the fact that $S^1\approx \Bbb R/\Bbb Z$, then the equivalence relations are $(x,y)\sim(x,y+n)$ for $n\in\Bbb Z$ (given by the circle) and $(x,y)\sim (-x,y+1/2)$. Combining these two pieces, we see that the equivalence relation is defined by $(x,y)\sim((-1)^nx,y+n/2)$. Therefore, $X=(S^2\times\Bbb R)/\Bbb Z$, where the action of $\Bbb Z$ is generated by $(x,y)\mapsto(-x,y+1/2)$.
Note that this group action is free, so we have $\pi_1(X)\cong\Bbb Z$. According to Hurewicz, $H_1(X;\Bbb Z)\cong \Bbb Z$ as well. Since this $\Bbb Z_2$-action on $S^2\times S^1$ is orientation reversing, $X$ is non-orientable, so $H_3(X;\Bbb Z)\cong 0$ and the torsion of $H_2(X;\Bbb Z)$ is $\Bbb Z_2$.
To determine the rank of $H_2$, we use the fact that the quotient map $q:S^2\times S^1\to X$ is a double cover, so $2\chi(X)=\chi(S^2\times S^1)=0\implies \chi(X)=0$. On the other hand, $$\chi(X)=b_0-b_1+b_2-b_3=1-1+b_2-0=b_2$$ where $b_i$'s are betti numbers. This shwos that the rank of $H_2$ is $0$.
In conclusion, $$H_k(X;\Bbb Z)\cong\begin{cases}\Bbb Z & k=0,1\\ \Bbb Z_2&k=2\\ 0 &\text{ otherwise}\end{cases}$$
Method 2: Cellular homology
The cell structure of $S^2$:
The cell structure of $S^1$ is constructed in the same way as $S^2$.
The reason to choose such non-standard cellular decomposition with 2-fold symmetry is that we want to obtain another cell structure when we descends to the quotient space $X$.
The cell structure $S^2\times S^1$ should be obvious from the construction above, i.e., four $3$-cells, eight $2$-cells, eight $1$-cells, and four $0$-cells. Now, consider the equivalence relation $(x,z)\sim (-x,-z)$, we see that cells are collapsed in pairs. For instance, $e_\alpha^1\times f_N^0$ is identified with $e_\beta^1 \times f_S^0$, so we have the following cellular chain complex for $X$.
$$0\to\Bbb Z^2\overset{\partial_3}{\to}\Bbb Z^4\overset{\partial_2}{\to}\Bbb Z^4\overset{\partial_1}\to\Bbb Z^2\to 0$$
Consider the one $e_\alpha^1\times f_N^0$, its image under the boundary map is $\partial_1(e_\alpha^1)\times f_N^0=(e_N^1-e_S^1)\times f_N^0$. We don't have to consider its "partner" $e_\beta^1\times f_S^0$ because they are identified in the quotient space. Similarly, one can argue that $\partial_1(e_\beta^1\times f_N^0)=(e_S^0-e_N^0)\times f_N^0=-\partial_1(e_\alpha^1\times f_N^0)$ and that $\partial_1(e_N^0\times f_\alpha^1)=e_N^0\times(f_N^0-f_S^0)=-\partial_1(e_N^0\times f_\beta^0)$. This shows that
Similar arguments can be repeated for $\partial_2$ and $\partial_3$ using the formula of cellular boundary map. Note that we need to refer back to the equivalence relation to see the relationship between $\ker\partial_{k}$ and $\operatorname{im}\partial_{k+1}$.
If we compute the quotient groups carefully, then we actually get the same answer as using method 1.