The mapping $z \mapsto \dfrac{1+z}{1-z}$ maps the full unit disc conformally onto the right half plane. You can check that the interval $(-1,1)$ maps onto the real axis, so it takes the upper semi-disc onto a quadrant. (You can verify that the image is the first quadrant.)
Then continue with the mapping $z \mapsto z^2$, taking the first quadrant conformally onto the upper half plane, and finish off with a Möbius map taking the upper half-plane back to the unit disc.
It depends on what is meant by "polynomial".
If only $\sum c_n z^n$, then every function that is uniformly approximable by polynomials must be holomorphic on the interior of $J$.
Although that condition is trivially satisfied if $J$ has empty interior, that doesn't mean that for such $J$ every continuous function is the uniform limit of polynomials. For example the unit circle has empty interior, but a sequence of polynomials converging uniformly on the unit circle converges uniformly on the closed unit disk by the maximum principle, and thus if $f$ is a uniform limit of polynomials on the unit circle, then there is a holomorphic function $h$ on the unit disk that extends continuously to the unit circle, with boundary values $f$. In particular, we have
$$\int_{\lvert z\rvert = 1} f(z)\cdot z^n \,dz = 0\tag{1}$$
for all $n \geqslant 0$. (And, in this case, that condition is sufficient.)
That phenomenon generalises, if $J$ disconnects the plane, that is, if $\mathbb{C}\setminus J$ has at least two connected components, then the bounded components of the complement of $J$ impose restrictive conditions on the continuous functions that are uniform limits of polynomials similar to $(1)$.
Mergelyan's theorem asserts the converse, if $J$ is a compact subset of $\mathbb{C}$ with empty interior such that $\mathbb{C}\setminus J$ is connected, then every continuous function on $J$ can be uniformly approximated by polynomials (in $z$ only).
If "polynomial" means polynomial in $z$ and $\overline{z}$, or equivalently polynomial in $\operatorname{Re} z$ and $\operatorname{Im} z$, then the Weierstraß approximation theorem holds for all compact $J$.
Best Answer
This is true for the circle (one needs Caratheodory's kernel theorem to prove it), but not in general. To see why, let $A:=\mathbb{D}-[0,1)$ open and simply connected (since its complement is connected). There exists a conformal map $\varphi: \mathbb{D}\to A$ and it admits a continuous extension to $\overline{\mathbb{D}}\to \overline{A}$ by Carathéodory-Torhorst. Note in particular that $\varphi^{-1}$ is bounded.
Define now $f$ as a Blaschke product (thus bounded) having $X=\cup\{(1-1/n^2)e^{iq_n}\}$ as zeros (where $q_n$ is an enumeration of the rationals in $[0,2\pi)$). This function is analytic bounded on $\mathbb{D}$ and non-extendible. Since $\varphi$ has a continuous extension to $\overline A$, $\overline{\varphi(X)} \supseteq\varphi(\overline{X})\supset \varphi(\partial\mathbb{D})=\partial A$.
Now, let $g=f\circ \varphi^{-1}$, which is clearly bounded and holomorphic on $A$. Suppose thre exists a sequence of bounded polynomials $p_n\to f\circ\varphi$. By continuity, $\{p_n\}$ must be bounded on $\overline{A}=\overline{\mathbb{D}}$. Montel's theorem implies the existence of a converging subsequence $p_{n_k}$. The limit must be an extension of $f\circ\varphi^{-1}$, and it must be analytic on $\mathbb{D}$. Since $[0,1)\subset \overline{g^{-1}(0)}$, the identity principle implies $g\equiv 0$, a contradiction.
This is true in general: a bounded function on an open connected set $A$ is the limit of a bounded sequence of polynomials only if it admits a bounded holomorphic extension to $\hat{\mathbb{C}}- \overline{(\hat{\mathbb{C}}-\overline{A})}$. This is actually a iff but the other direction of the implication is longer.