Can anyone provide me with an example of two non-homeomorphic locally-compact Hausdorff spaces $X$ and $Y$, such that $C(X)$ and $C(Y)$ are isomorphic as Banach algebras. Clearly, the Gelfand--Naimark 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 1-point 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.