The short answer is that your presentation is isomorphic to the (slightly more) "standard" presentation - so you computed a correct answer.
More detail -
According to Wolfram and using $K$ to denote the Klein bottle, we have $\pi_1(K) \cong \langle c,d \rangle/cdc^{-1}d$ whereas you have (according to Aaron Mazel-Gee's comment) $\pi_1(K)\cong \langle a,b\rangle /abab^{-1}$.
The question is, then, are these isomorphic?
Well, we have $cdc^{-1}d = e$ so, taking the inverse of both sides gives $d^{-1}cd^{-1}c^{-1} = e$. But this has the same form as the relation between $a$ and $b$, so now the isomorphism is clear: we map $a$ to $d^{-1}$ and $b$ to $c$.
That is, define $f:\langle a,b \rangle /abab^{-1}\rightarrow \langle c,d\rangle/cdc^{-1}d$ by $f(a) = d^{-1}$ and $f(b) = c$.
I claim this is well defined, for \begin{align*} f(abab^{-1}) &= f(a)f(b)f(a)f(b)^{-1} \\\ &= d^{-1}cd^{-1}c^{-1} \\\ &= (cdc^{-1}d)^{-1} \end{align*} as it should. (Technically, I'm defining $f$ on $\langle a,b\rangle$ and proving it descends to the quotient.)
Finally, rather than show this is 1-1 and onto, instead, show that $g:\langle c,d\rangle/ cdc^{-1}d\rightarrow \langle a,b\rangle/ abab^{-1}$ defined by $g(c) = b$ and $g(d) = a^{-1}$ is well defined, and the inverse to $f$, so $f$ is the desired isomorphism.
One subgroup of $\langle a, b\ |\ abab^{-1} = 1 \rangle $ that is isomorphic to $\mathbb{Z}\times \mathbb{Z}$ is the subgroup generated by $a$ and $b^2$. Indeed these elements commute: $ab^2 = ba^{-1}b = b^2 a$, and as this is a finite-index sub-group of a surface group it is also a surface group, so in particular it only has one relation. In general you could take $a^m$ and $b^{2n}$.
In an answer to a related question (Is there a non-trivial covering of the Klein bottle by the Klein bottle?) I gave a few families of subgroups and determined the total spaces of the corresponding cover space in those cases, maybe you will find it useful.
Best Answer
The concept you're probably looking for is the deck transformation group of a covering space. In this case, your group $G$ happens to be an invariant composite of the covering $\phi$, that is, for all $\rho\in G$, the composition $\phi\circ\rho=\phi$. It follows that, as $\mathbb{R}^2/G$ is the full orbit space, it must be that $G$ is the full deck transformation group for the covering $\phi$.
Now, $\mathbb{R}^2$ is a simply connected topological space (in fact it's contractible) and so it is not just a cover but a universal cover of the Klein bottle. Universal covers have the nice property of always being regular coverings (the induced action of the deck transformation group on the fiber of the cover is both free and transitive) and so the deck transformation group $G$ is isomorphic to the fundamental group $\pi_1(\mathbb{R}^2/G)$ of the Klein bottle.
The justification of the last claim is easiest to describe using the theory of principle $G$-bundles although this might be a bit out of the scope of your current course so feel free to ignore this if you get lost (I believe Hatcher gives a very good elementary explanation of the isomorphism without reference to bundles, so feel free to check his textbook). For completeness though, we'll show that $G\cong\pi_1(\mathbb{R}^2/G)$. Let $F$ be the fiber of the map $\phi$. The principle $G$-bundle $\phi$ induces an exact sequence in homotopy $$\pi_1(\mathbb{R}^2)\stackrel{\phi_*}{\longrightarrow}\pi_1(\mathbb{R}^2/G)\longrightarrow\pi_0(F)\longrightarrow\pi_0(\mathbb{R}^2).$$ Now $\pi_1(\mathbb{R}^2)$ is trivial, and $\mathbb{R}^2$ is path connected so $\pi_0(\mathbb{R}^2)$ is trivial. We know that $\pi_0(F)$ is actually a group and is isomorphic to $G$ because $\phi$ is a principle $G$-bundle. It follows that we have the short exact sequence $$0\stackrel{\phi_*}{\longrightarrow}\pi_1(\mathbb{R}^2/G)\longrightarrow G\longrightarrow 0$$ and so we have an induced isomorphism $\pi_1(\mathbb{R}^2/G)\stackrel{\cong}{\longrightarrow} G$.