Is the isomorphism problem for amenable groups decidable? Is it algorithmically decidable if two finitely presented amenable groups are isomorphic?
Or slightly different:
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?
 A: 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.
For example, a similar argument shows that there is an enumeration $G_0,G_1,G_2,\ldots$ of copies of the two groups $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, such that the isomorphism problem is not decidable. Simply make $G_n\cong\mathbb{Z}$ if $n\in A$ and otherwise $G_n\cong \mathbb{Z}/2\mathbb{Z}$, for a fixed undecidable set $A$. 
Of course this isn't the answer you or anyone is interested in,
even though it does 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. 
A: EDIT: The isomorphism problem for finitely presented solvable groups in the variety of all solvable groups of derived length $\le 7$ is undecidable. This was proved by Kirkinskiĭ and Remeslennikov (Kirkinskiĭ, A. S.; Remeslennikov, V. N.
`The isomorphism problem for solvable groups.' (Russian)
Mat. Zametki 18 (1975), no. 3, 437–443.). The Russian version of this article can be downloaded from here. The English translation is available here.
Unfortunately this does not fully answer the original question, because the groups in this construction are finitely presented in the variety of solvable groups but may not be finitely presented in the variety of all groups. 
I would guess that one could use O. Kharlampovich's example of a finitely presented 3-step solvable group with unsolvable word problem (Harlampovič, O. G., `A finitely presented solvable group with unsolvable word problem.' (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 45 (1981), no. 4, 852–873, 928.) to construct the family of groups you need. Perhaps someone has already done this... 
Second EDIT: Indeed, this was done by Baumslag, Gildenhuys and Strebel (see Theorem 1 in Baumslag, Gilbert; Gildenhuys, Dion; Strebel, Ralph, `Algorithmically insoluble problems about finitely presented solvable groups, Lie and associative algebras. II.'
J. Algebra 97 (1985), no. 1, 278–285.), who proved that the isomorphism problem is undecidable in the class of finitely presented solvable groups of derived length 3.
In fact, in one of her talks Olga Kharlampovich mentioned that she can construct a finitely presented 3-step solvable group $G$ with unsolvable word problem that is Hopfian. Then the isomorphism problem among the quotients of $G$ by one defining relation is unsolvable (because $G/\langle\langle  g \rangle\rangle^G$ is isomorphic to $G$ if and only if $g=1$ in $G$).
A: According to the diagram on page 31 of Chuck Miller's authoritative MSRI notes, the isomorphism problem is known to be undecidable for finitely presented solvable groups of derived length 3, and hence for amenable group.  I'll add further details when I have time.
