This answer is to point out that the particular way in which you
have posed the second question is not actually the question you
meant to ask, for it admits a trivial answer, requiring almost no knowledge about amenable groups. Namely, you ask,

>> Does there exist a family of amenable groups (indexed by natural numbers) for which one cannot algorithmically decide if two elements of the family are isomorphic?

The answer is yes. Fix any non-computable set
$A\subset\mathbb{N}$, and then enumerate the groups you are
interested in $G_0,G_1,G_2,\ldots$ (assuming there are at least
two non-isomorphic such groups) in such a way so that the indices
$n$ of one of the isomorphism classes is exactly $A$, and use the
rest of the indices for the rest of the groups (or just some of
them) in an arbitrary manner. That is, we make all $G_n$ for $n\in
A$ isomorphic, and not isomorphic to any other $G_m$ for $m\notin
A$. With such an enumeration, the isomorphism problem is not
decidable, simply because for a fixed $k\in A$, we cannot tell if
$G_n\cong G_k$, because this would provide a decision procedure
for $n\in A$, which is undecidable.

Of course this isn't the answer you or anyone is interested in,
even though it does actually answer the question you actually
asked. So the point is that you shouldn't consider such arbitrary
enumerations when asking decidability questions about finitely
presented groups, but rather you want to ask about decidability
questions for the more natural enumerations of the presentations
that arise from a natural indexing of the presentations, by means
of a coding of the syntax of the presentation.