I want to compute the presentation for the kernel of $f:G\to(\mathbb{Z},+)$ given by $f(a)=f(b)=0$ and $f(c)=1$, where $G=\langle a,b,c\mid aba=bab,bc=cb\rangle$. It is easy to see that $a^{c^i}:=c^iac^{-i}\in\ker(f)~\forall~i\in\mathbb{Z}$, so we can expect to have $\ker(f)$ generated by $\lbrace b,a^{c^i}\rbrace_{i\in\mathbb{Z}}$. Hence, I claim that the kernel is: $$\langle b,a^{c^{i}},i\in\mathbb{Z}\mid a^{c^{i}}ba^{c^{i}}=ba^{c^{i}}b,i\in\mathbb{Z}\rangle$$
But I dont know whether this is right or not. It seems obvious that the kernel is generated by those elements, since every element $w\in\ker(f)$ I've tried can be obviously written as a product of those elements, however I can't prove it.
Any help will be appreciated.
As I said in my comments, you can do this using the Reidemeister-Schreier algorithm for computing presentations of subgroups of groups defined by a presentation. Let $K = \ker f$ be the subgroup.
To apply the algorithm, we first need a Schreier transversal $T$ of $K$ in $G$, which means a prefix-closed set of words in the group generators that map onto a set of (right) coset representatives of the subgroup. The obvious choice of $T$ in this case is $T = \{c^i : i \in {\mathbb Z} \}$. Formally, we define $c_0 := 1$ (the identity element) and define $c_{i+1} := c_ic$ for $i \ge 0$ and $c_{-i+1} = c_{-i}c^{-1}$ for $ i \ge 0$. Then of course $c_i$ is equal in $G$ to $c^i$.
Now $Kc_ia = Kc_i$ and $Kc_ib=Kc_i$ for all $i \in {\mathbb Z}$, so there are elements $a_i,b_i \in K$ with $c_ia = a_ic_i$ and $c_ib = b_ic_i$ for all $i \in {\mathbb Z}$, and by a well-known result of Schreier, we have $K = \langle a_i,b_i \mid i \in {\mathbb Z}\rangle$. As elements of $G$, we have $a_i = c^iac^{-i}$ and $b_i = c^ibc^{-i}$.
Now we can use the Reidemeister method to calculate a set of defining relations of $K$ on these generators. To do this we apply each relation to each $c_i$ on the right.
For the first relation $aba=bab$, $c_iaba = c_ibab$ yields the relation $a_ib_ia_i = b_ia_ib_i$ of $K$ for each $i$.
For the second relation $bc=cb$, $c_ibc = c_icb$ yields $b_i=b_{i+1}$ for all $i$ so, so after replacing each $b_i$ by $b_0$, we get $$ K \cong \langle a_i\,(i \in {\mathbb Z}),\,b_0 \mid a_ib_0a_i=b_0a_ib_0\,(i \in {\mathbb Z}) \rangle$$ and, since $a_i = c^iac^{-i}$ and $b_0=b$, this is equivalent to your conjectured presentation.
Or, if you want to be formal, the map $a_i \mapsto c^i ac^{-i}\,(i \in {\mathbb Z})$, $b_0 \mapsto b$ induces an isomorphism from group defined by the above presentation onto the subgroup $K = \ker f$ of $G$.