This is more of a long comment than an answer. We call a category $C$ concrete if it's equipped with a forgetful functor $U : C \to \text{Set}$, usually assumed to be faithful; this formalizes the intuitive notion of a category of "sets with extra structure," where $F$ describes the underlying set of an object. The property you want, that a morphism in $C$ which is bijective on underlying sets is an isomorphism, corresponds to $U$ being conservative. A conservative functor is one that reflects isomorphisms, meaning that if $F(f)$ is an isomorphism then $f$ is an isomorphism.
Faithful and conservative functors can be related as follows. First, some nonstandard definitions: say that a morphism is a pseudo-isomorphism if it is both a monomorphism and an epimorphism, and a fake isomorphism if it's a pseudo-isomorphism, but not an isomorphism.
Exercise 1a: Faithful functors reflect epimorphisms and monomorphisms: that is, if $F$ is faithful and $f$ is a morphism, then if $F(f)$ is an epimorphism then $f$ is an epimorphism, and if $F(f)$ is a monomorphism then $f$ is a monomorphism. Hence faithful functors reflect pseudo-isomorphisms.
Exercise 1b: If $F : C \to D$ is a faithful functor and $C$ has no fake isomorphisms (so every pseudo-isomorphism is an isomorphism), then $F$ is conservative.
Hence, if $C$ is a concrete category whose forgetful functor isn't conservative, then $C$ must have fake isomorphisms. $C = \text{Top}$ is a well-known example; in this category fake isomorphisms exist because we can add open sets to a topology and get another topology, which allows us to construct continuous bijections which are not homeomorphisms.
In addition, while it's not true in general that pseudo-isomorphisms are isomorphisms, there are many statements of the form "a morphism which is both a monomorphism and a (some special kind of epimorphism) is an isomorphism." A reasonably useful one in practice is:
Exercise 2a: A morphism which is both a monomorphism and an effective epimorphism is an isomorphism.
Exercise 2b: If $F : C \to D$ is a faithful functor and every epimorphism in $C$ is effective, then $F$ is conservative.
The condition that every epimorphism is effective holds in some categories of algebraic objects, such as $\text{Vect}$ and $\text{Grp}$, but not in others, such as $\text{Ring}$.
It turns out that in $\text{CHaus}$ every epimorphism is effective; what this says somewhat more concretely is that every continuous surjection $X \to Y$ between compact Hausdorff spaces is a quotient map, or in other words that $Y$ has the quotient topology (note that this is emphatically not true in $\text{Top}$!). so this is one way of explaining why $\text{CHaus}$ has a conservative forgetful functor. I don't think this is true in the category of Banach spaces though.
The comments allude to the fact that monadic functors are conservative, and while this covers the case of compact Hausdorff spaces it doesn't cover the case of Banach spaces.
$Y$ can be described as the Stone space of the measure algebra of $(X,\mu)$. That is, let $\Sigma$ be the $\sigma$-algebra on which $\mu$ is defined, let $N\subseteq\Sigma$ be the ideal of null sets, and let $B=\Sigma/N$ be the quotient Boolean algebra. Then $Y$ is naturally homeomorphic to the set $S$ of Boolean homomorphisms $B\to\{0,1\}$, topologized as a subspace of $\{0,1\}^B$.
To prove this, let us first recall that $Y$ can be described as the set of $*$-homomorphisms $L^\infty(X,\mu)\to\mathbb{C}$, with the topology of pointwise convergence. For each $b\in B$, there is a function $1_b\in L^\infty(X,\mu)$, and a $*$-homomorphism $\alpha$ must send $1_b$ to either $0$ or $1$ since $1_b^2=1_b$. It is then easy to see that $b\mapsto \alpha(1_b)$ is a Boolean homomorphism $B\to\{0,1\}$ (the Boolean operations on sets can be expressed in terms of ring operations on their characteristic functions). This defines a map $F:Y\to S$.
Note moreover that since simple functions are dense in $L^\infty(X,\mu)$, an element of $Y$ is determined by its values on characteristic functions $1_b$. Thus $F$ is injective. Also, $F$ is continuous, since the topology on $S$ is the topology of pointwise continuity with respect to evaluation at just the elements $1_b$. Since $Y$ and $S$ are both compact Hausdorff, it follows that $F$ is an embedding.
It remains to be shown that $F$ is surjective. Fix a homomorphism $h:B\to\{0,1\}$, and let $U=h^{-1}(\{1\})$. The idea is that we can then define a $*$-homomorphism $L^\infty(X,\mu)\to\mathbb{C}$ which maps a function $f$ to the "limit" of the values of $f$ along the ultrafilter $U$. To make this precise, given $f\in L^\infty(X,\mu)$ and $b\in B$, let $f[b]\subset\mathbb{C}$ denote the essential range of $f$ on $b$, and let $C_f=\{f[b]:b\in U\}$. Note that each element of $C_f$ is compact and nonempty. Also, $f[b\cap c]\subseteq f[b]\cap f[c]$, so $C_f$ has the finite intersection property. Thus $\bigcap C_f$ is nonempty. If $x\in \bigcap C_f$, then for any neighborhood $V$ of $x$ and any $b\in U$, $f^{-1}(V)\cap b$ is non-null. Since $U$ is an ultrafilter on $B$, this means $f^{-1}(V)\in U$. Now if we had two different points $x,y\in C_f$, they would have disjoint neighbooods $V$ and $W$, and then $f^{-1}(V)$ and $f^{-1}(W)$ would be disjoint elements of $U$. This is impossible.
Thus, we have shown that $C_f$ has exactly one point for each $f\in L^\infty(X,\mu)$. Define $\alpha(f)$ to be the unique element of $C_f$, which can also be described as the unique point $x$ such that given any neighborhood $V$ of $x$, for all sufficiently small $b\in U$, $f|_b$ takes values in $V$ almost everywhere. This description makes it easy to verify that $\alpha$ is a $*$-homomorphism, and that $\alpha(1_b)=h(b)$ for each $b\in B$. Thus $\alpha\in Y$ and $h=F(\alpha)$, so $h$ is in the image of $F$, as desired.
(Alternatively, to show $F$ is surjective, by Stone duality it suffices to show that the image of $F$ separates elements of $B$, since closed subspaces of the Stone space $S$ correspond to quotients of the algebra $B$. But by Gelfand duality, elements of $Y$ separate elements of $L^\infty(X,\mu)$, and so we're done since distinct elements of $B$ have distinct characteristic functions in $L^\infty(X,\mu)$.)
Best Answer
It is true that the category of locally compact Hausdorff spaces is equivalent to the category of commutative $C^*$-algebras . . . with appropriately chosen morphisms.
Let $A$ and $B$ be commutative $C^*$-algebras. Then, a morphism from $A$ to $B$ is defined to be a nondegenerate homomorphism of $^*$-algebras from $\phi :A\rightarrow M(B)$, where $M(B)$ is the multiplier algebra of $B$. Here, nondegeneracy means that the the span of $\left\{ \pi (a)b:a\in A,b\in M(B)\right\}$ is dense in $M(B)$. Note that you need a bit of machinery to even make this into a category because it is not obvious a priori that composition makes sense. Nevertheless, it does work out. Proposition 1 on pg. 11 and Theorem 2 on pg. 12 of Superstrings, Geometry, Topology, and $C^*$-algebras (in fact this chapter is on the arXiv) respectively show that this forms a category and that the dual of this category is equivalent to the category of locally compact Hausdorff spaces.