Can anyone provide me with an example of two nonhomeomorphic locallycompact Hausdorff spaces $X$ and $Y$, such that $C(X)$ and $C(Y)$ are isomorphic as Banach algebras. Clearly, the GelfandNaimark theorem tells us that the two algebras $C(X)$ and $C(Y)$ cannot be isomorphic as $C^*$algebras.
There are none.
To be more precise, for locally compact spaces $X$ you have to consider $C_0(X)$, the space of continuous functions which vanish at $\infty$; you can reduce this case to that of compact $X$ by considering the 1point compactification and removing the constants by $f\mapsto f f(\infty).1$.
Now, for compact $X$, the Banach algebra $C(X,\mathbb C)$ is always a commutative $C^*$algebra, thus $X$ is isomorphic to the closed subspace of multiplicative $C^*$functionals in the duals. Thus your question has a negative answer.
For real valued functions, you have to use the Banach lattice property and the condition $f,g\ge0\implies \f\vee g\=\f\\vee\g\$ to get the same result (Stone's theorem).
But consider the following Banach lattice (see Semadeni: Banach spaces of continuous functions, 1971): Let $U$ be be a bounded domain in $\mathbb R^n$. Then the space $H_b(U)$ of bounded harmonic functions on $U$ has the above property, but the Banach lattice supremum is not the pointwise supremum. In fact, $H_b(U)\cong L^\infty(\partial U)$. I hope that this catches what you were looking for.

$\begingroup$ Hi Peter. Thanks a lot for your answer! Yes this answers my question. What I was wondering was whether the existence of the $\ast$structure was necessary for differentiating l.c.H.sps using their cts function algs  your answer tells me that no it is not. So the point of GelfandNaimark is that one needs the $*$structure in order to identify those Banach algebras that admit a $*$structure? Moreover, when this $*$structure exists it must then be unique, no? $\endgroup$ – Janos Erdmann Oct 20 '13 at 16:15