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
Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$\langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2\rangle$$ which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)