Small-cancellation theory

At the OP's request, here are some references for the last sentence; they are all from Lyndon and Schupp's book Combinatorial group theory.

Take $r_1,r_2,\ldots$ to be an infinite sequence of elements of $F(b_1,b_2)$ such that, for every $n$, $R_n={r_1,\ldots,r_n}$ satisfies condition $C'(1/6)$ (as defined on p. 240 of Lyndon and Schupp). It's a nice exercise to confirm that such sequences exist.

For each $n$, take $G_n=A*\langle b_1,b_2\mid R_n\rangle$. It's easy to check that there are infinitely many elements $g_i$ so that, for all distinct $i,j$, $g_ig_j^{-1}$ is $R$-reduced in the sense of p. 251 of Lyndon ad Schupp. Therefore, by Dehn's algorithm, $G_n$ is infinite for all $n$, as claimed.

Equational Noetherian groups

In response to a remark of the OP's in the comments, I want to point out that it does not follow that $G_0$, or any $G_n$, is not equationally Noetherian. Indeed, the $G_n$ constructed above are all word-hyperbolic and hence equationally Noetherian by a theorem of Sela. (Alternatively, for $C'(\lambda)$ for small enough $\lambda$, they are all linear by the work of Wise and friends, and hence are equationally Noetherian by Hilbert's Basis Theorem.)

To prove that $G_0$ (say) is not equationally Noetherian, you need an infinite sequence of proper epimorphisms

$L_0\to L_1\to L_2\to\cdots$

where each $L_n$ is residually $G_0$.

For more details, I suggest you look at the beginning of Bestvina and Feighn's paper Notes on Sela's work (here).

2 Corrected typo.

The answer to your question is 'no'. Consider any homomorphism $f:A*F\to G$ which is injective on $A$. Then $\ker f$ is an ideal in your sense (and this is necessary and sufficient). An infinite increasing chain of ideals is therefore equivalent to an infinite sequence of surjections

$A*F=G_0\to G_1\to G_2\to\ldots$

so that $A$ embeds into $G_n$ for all $n$. For a specific example, take $F=\langle b_1,b_2\rangle$ and construct $G_n$ from $G_{n+1}$ G_{n-1}$by adding a long small-cancellation relator in$b_1,b_2$. 1 The answer to your question is 'no'. Consider any homomorphism$f:A*F\to G$which is injective on$A$. Then$\ker f$is an ideal in your sense (and this is necessary and sufficient). An infinite increasing chain of ideals is therefore equivalent to an infinite sequence of surjections$A*F=G_0\to G_1\to G_2\to\ldots $so that$A$embeds into$G_n$for all$n$. For a specific example, take$F=\langle b_1,b_2\rangle$and construct$G_n$from$G_{n+1}$by adding a long small-cancellation relator in$b_1,b_2\$.