Does there exist a noetherian domain $A$ and a principal ideal $I = (x)$ in it having an embedded component?
Let $X$ be a locally Noetherian scheme, e.g., we have that $X =$ Spec $A$, where $A$ is a Noetherian ring.
Then, if $X$ is reduced (as a scheme), then $X$ has no embedded points.
In particular, if $A$ is a domain, then $X$ has no embedded points.
The converse isn't true. As you probably know, the scheme Spec $k[u,v]/(u^2,uv)$, where $k$ is a field, admits an embedded point corresponding to the maximal ideal $(u,v)$.
Maybe I'm misunderstanding the question?
Take the homogeneous coordinate ring of the rational quartic curve, that is, $$R=K[a,b,c,d]/(ad-bc, a^2c-b^3, bd^2-c^3, ac^2-b^2d).$$ (Note that this ring is isomorphic to the affine semigroup ring $K[x^4,x^3y,xy^3,y^4]$ and now we have a connection to the answer from this topic.) Then chose any nonzero noninvertible homogeneous element of $R$ and get the required example.
If you want to study more I can recommend you this paper.
Remark. In the book of B. Hassett, Introduction to Algebraic Geometry, exercise 8.9 provides an example of a principal ideal that has embedded associated primes. His example looks pretty similar to the one given above, but isn't an integral domain. (I wonder whether Hassett missed somehow one generator from the defining ideal of his ring $R$, because for rings which are not integral domains there are much simpler examples.)