Here is an example (the group $H_K^\mathbf{Q}$ defined below, which is actually a split extension abelian-by-countable).
Let $(Q,\le)$ be a total ordering. Let $K$ be a countable (possibly finite) field. Let $V^Q_K=K^{(Q)}$ be the free $K$-module with basis $(e_q)_{q\in Q}$. Let $G^Q_K$ be the group of $K$-module automorphisms $g$ of $V_K$ such that $g(e_q)-e_q\in\mathrm{Vect}_Q(\{e_r:r<q\})$ for every $q$ (it can be thought as a upper unipotent group of $Q\times Q$-matrices over $K$).
Define a cut as a subset $I$ of $Q$ such that both $I$ and its complement $I^c$ are nonempty, and such that $i<j$ for all $i\in I$ and $j\in I^c$. For every cut $I$ with complement $J=I^c$ and $g\in G^Q_K$, let $p_I(g):V^I_K\to V^I_K$ be its restriction to the stable subspace $V^I_K$, and let $r_I(g):V^J_K\to V^J_K$ be the induced automorphism modulo the stable subspace $V^I_K$. (Visually, $I$ defines a $2\times 2$ block decomposition, $p_I(g)$ is the northwest block and $r_I(g)$ is the southeast block of $g$.)
Obviously, $p_I$ and $r_I$ are homomorphisms, and if $P_I$ and $R_I$ are their kernels, then $N_I=P_I\cap R_I$ is is a normal abelian subgroup in $G^Q_K$.
Now we specify to $Q=\mathbf{Q}$, the rationals with the usual ordering.
Still, $N_I$ is not of countable index in $G^\mathbf{Q}_K$. So we define $H^Q_K$ as the normal subgroup of $G^Q_K$ of elements $g$ such that for every cut $I$, both matrices $p_I(g)$ and $r_I(g)$ are finitely supported (i.e. finitely many non-diagonal elements are nonzero). Then $N_I\cap H^\mathbf{Q}_K$ is normal abelian of countable index in $H_K^\mathbf{Q}$.
It remains to see that $H_K^\mathbf{Q}$ has no abelian characteristic subgroup of countable index. This is rather easy: the first observation (an easy play with commutators(*)) is that every nontrivial normal subgroup $N$ of $H_K^\mathbf{Q}$ contains an elementary matrix $e_{ij}(s)$ for some $i<j$ and $s\neq 0$. All such matrices are conjugate under the full automorphism group of $H_K^\mathbf{Q}$ (to vary $s$, use diagonal matrices, and to vary $(i,j)$, use that the group of order-preserving permutations of $\mathbf{Q}$ is 2-transitive). Hence if $N$ is characteristic, it contains all elementary matrices, and thus the (non-abelian) subgroup they generate.
Hence the only abelian characteristic subgroup of $H_K^\mathbf{Q}$ is the trivial subgroup $\{1\}$: we have to check that it does not have countable index, i.e. $H_K^\mathbf{Q}$ is uncountable: indeed for every subset $L$ of $\mathbf{N}\smallsetminus\{0\}$, consider the matrix $k_L$ with same entries as the identity except $(k_L)_{-i,i}=1$ for all $i\in L$. Then $L\mapsto k_L$ is injective, $k_L$ belongs to $H_K^\mathbf{Q}$ for every $L$, hence $H_K^\mathbf{Q}$ has continuum cardinal.
Edits:
(1) (*) Starting from a non-identity element $g$ in $G_K^\mathbf{Q}$, its commutator with a suitable elementary matrix yields a non-identity element $h$ such that $h-1$ has finite rank; in particular if $g\in H_K^\mathbf{Q}$ then so does $h$, and it follows that $h$ is finitely supported. Then $h$ lies in the nilpotent subgroup $G_K^I$ for some finite subset $I$ of $\mathbf{Q}$. So the normal subgroup of $H_K^\mathbf{Q}$ generated by $g$ contains a central element of $G_K^I$, showing that any nontrivial normal subgroup of $H_K^\mathbf{Q}$ contains a non-identity elementary matrix.
(2) Actually, a similar argument shows that the only normal abelian subgroup of $H_K^\mathbf{Z}$ stable under the shift automorphism $\alpha$ (conjugation by the automorphism $(e_n\mapsto e_{n+1})_{n\in\mathbf{Z}}$) is the trivial subgroup $\{1\}$. Note that $H_K^\mathbf{Z}$ is residually nilpotent, while $H_K^\mathbf{Q}$ is a perfect group.
(3) Incidentally, with the notation of (2), the semidirect product $\langle\alpha\rangle\ltimes H_K^\mathbf{Z}$ has a countable index abelian subgroup but no normal one. I don't know whether the Podoski-Szegedy example is similar (I don't have access to their article right now).
