Is the Fundamental Groupoid a Bicategory?

It isn't a bicategory, but this doesn't have to do with $2$-cells being isomorphisms. (They're really equivalences and, in fact, the $1$-cells are also equivalences.) The reason it fails to be a bicategory is that composition of $2$-cells isn't strictly associative or unital: in fact, this is exactly the same problem that means that points-and-paths doesn't define a category.

It sounds like what you are trying to describe is the fundamental ∞-groupoid $\pi(X)$ of $X$, which has the structure of an ∞-category (in fact, a (∞,0)-category), rather than a bicategory. The $n$-cells are the "$n$-dimensional paths" in $X$, and the composition axioms hold only up to equivalence.

The fundamental groupoid $\pi_1(X)$ is then the $1$-truncation of $\pi(X)$, which quotients the $1$-cells (i.e. paths) by higher equivalence. The resulting structure is a groupoid, which is in turn a category.

The structure you (almost) describe is the $2$-truncation of $\pi(X)$, which is indeed a bicategory (in fact, it's a $2$-groupoid), except the $2$-cells are equivalence classes of homotopies, rather than homotopies themselves. This ensures that the rules governing composition of $2$-cells hold strictly.

More generally, the $n$-truncation $\pi_n(X)$ of $\pi(X)$ has points as 0-cells, paths as $1$-cells, homotopies as $2$-cells, homotopies of homotopies as 3-cells, and so on. When you get to the level of $n$-cells, you quotient by higher homotopy to ensure the composition laws for $n$-cells hold strictly; the obtained structure is an $n$-groupoid, a type of weak $n$-category, whatever that means.

The other answer talks about the fundamental $2$-groupoid. I'll discuss the other possible correction to your construction: don't distinguish between parallel $2$-arrows.

Rather than having the $2$-morphisms be homotopies between homotopies, you can just have $2$-morphisms give the "is homotopic to" relation. That is, given paths $f$ and $g$, $\hom(f,g)$ is a one-point set if $f \simeq g$, and is empty otherwise.

The obvious functor from this construction to the fundamental groupoid is an equivalence, since for points $P$ and $Q$, the category $\hom(P,Q)$ is equivalent to the set of homotopy classes of paths from $P$ to $Q$.