You can compute the normal closure by computing the quotient, and then considering the kernel of the quotient homomorphism.
For the example you gave, let $N$ be the normal closure of $y$ in $G$. Then $G/N$ has presentation
$$
\langle x,y \mid x^{12}y = yx^{18},y=1\rangle
$$
This presentation reduces to $\langle x \mid x^{12} = x^{18}\rangle$, which is the same as $\langle x \mid x^6 = 1\rangle$.
Thus $G/N$ is a cyclic group of order 6, and $N$ is the kernel of the homomorphism $G\to G/N$. In particular, the normal closure of $y$ consists of all words for which the total power of $x$ is a multiple of 6.