The question "Is there a nonzero homomorphism from your group to $A_5$?" is decidable. (Just write down all ways of sending the generators of your group to $A_5$, and see whether they satisfy the required relations.) The same is true with $A_5$ replaced by any finite group. I don't see how to reduce this to questions about the abelianization.
This is an answer for the abelian case only.
For a finite abelian group (let us exclude the trivial one) there are two standard ways to decompose it as a direct sum of cyclic groups. One into cyclic groups of order $n_1,\dots,n_r$ (not $1$) such that $n_i \mid n_{i+1}$ and the other one into cyclic groups of order $m_1, \dots, m_s$ such that each $m_i$ is a prime power (not $1$). [Under each of the assumptions 'divisibility' and 'prime power' the repective decomposition is unique, in the latter case of course up to the ordering.]
The parameter $r$ is sometimes called the rank and the parameter $s$ the total rank of the group (although terminology here is not completely uniform).
Now, it is known that the rank is the minimal cardinality of a generating set, in the sense that there does not exists a set of a smaller cardinality that generates the group.
And, that the total rank is the maximal cardinality of a minimal generating set, that is there exists a generating set of cardinailty $s$ such that no subset of this set generates the group.
Thus, the cardinality of all minimal/irredundant generating set is uniquely determined if and only if the rank equals the total rank.
This is the case if and only if the group is an (abelian) $p$-groups.
Note that for a non-$p$-group on can first consider the first decomposition into cycylic groups, and then decompose each cyclic component as the sum of cyclic groups of prime power order (more or less Chines Remainder Theorem); so only if all the orders in the first decomposition are also prime powers (and thus necessarily powers of the same prime) does one not get a different value for rank and total rank.
Best Answer
Just to expand my comment, a Tarski Monster is an infinite group in which, for some fixed prime $p$, all proper nontrivial subgroups have order $p$. It was proved by Olshanskii in 1979 that they exist for all primes $p>10^{75}$.
A set of $p+1$ distinct elements of such a group cannot all lie in the same subgroup of order $p$, so they must generate the whole group, and so we have an upper bound of $p+1$ on the size of minimal generating sets.
It would interesting to know whether there are any less exotic examples.