I'm curious to know where you came across this question - it is the kind of thing people (well, at least one person) like to set as an exercise when giving a graduate course on Banach algebras.

Anyway: an alternative way to see that $A(U)$ is not algebra-isomorphic to the disc algebra is to look at their groups of invertible elements.

Every invertible element of the disc algebra has a logarithm inside the algebra: that is, if $f\in A(\overline{\mathbb D})$ is invertible, then $f=e^g$ for some $g\in A(\overline{\mathbb D})$. (In particular, the group of invertible elements in $A({\mathbb D})$ is path-connected.)

On the other hand, consider the following element of $A(\overline{U})$: let $f(z)=z$. Clearly this is invertible in $A(\overline{U})$. On the other hand, if $g\in A(\overline{U})$ and $f=e^g$ then the restriction of $g$ to the circle of radius $3/2$ would be a continuous single-valued logarithm on that circle, which is impossible for topological reasons.
(Pushing this argument further, one finds that the group of invertible elements in $A(\overline{U})$ is not connected.)

For more on this theme, look up the "Arens-Royden theorem". Having said all this, Faisal's answer is by far the simpler and neater way to approach the problem.