I am now very dubious that one can deduce anything about the sign of $L(\sigma,\chi)$ for $0<\sigma<1$ from the positivity of
$$
\sum_{n<x}\frac{\chi(n)}{n}(1-n/x)^k.
$$
Here's why. For simplicity consider the case $k=0$, and $\chi$ odd, i.e. complex quadratic. Bateman and Chowla (in the article M&V cite) point out that $L(1,\chi)>d^{-1/2}$, while Summation by Parts and the Polya-Vinograov inequality bounds the tail of the infinite series by
$$
\sum_{x<n}\frac{\chi(n)}{n}<\frac{5}{3}\cdot \frac{d^{1/2}\log(d)}{x}.
$$
Thus
$$
\sum_{n<x}\frac{\chi(n)}{n}>d^{-1/2}-\frac{5}{3}\cdot \frac{d^{1/2}\log(d)}{x}.
$$
The above is positive for
$$
x>\frac53 d\log(d),\quad\text{or}\quad d<\frac{3x}{5W(3x/5)},
$$
where $W(x)$ is the Lambert function, that is, the inverse function of $x=w\exp(w)$. Thus assuming one could show that
$$
\sum_{n<x}\frac{\lambda(n)}{n}>0
$$
for some very large $x$ (Haselgrove's disproof of Turan's conjecture suggests that $x=\exp(853)$ might be possible), one would be able (if the premise of the problem were correct) to rule out the possibility of an exceptional zero for a very large collection of $d$. For example, with $x=\exp(853)$ one would get all $d<2\cdot 10^{367}.$
A similar argument with fixed $k>0$ would give still more $d$.
Let's note that this is not a question of whether Collatz is undecidable.
The statement $\neg\mathrm{Con}(PA)$ is undecidable (by $PA$, assuming $PA$ is consistent) but nevertheless $\neg\mathrm{Con}(PA)$ is provably equivalent to a certain Turing machine halting (the one that searches for a proof of a contradiction in PA).
Rather, the question is whether
there is a $\Pi^0_1$ statement $\varphi$ such that the Collatz problem, which on its face is $\Pi^0_2$, is already known to be equivalent to $\varphi$.
Here already known means in particular that we are not allowed to assume that Collatz is or is not provable or disprovable in any particular system, unless we already know that.
The best evidence that there is no such $\varphi$ seems to be in the paper mentioned by @Burak:
Kurtz, Stuart A.; Simon, Janos, The undecidability of the generalized Collatz problem, Cai, Jin-Yi (ed.) et al., Theory and applications of models of computation. 4th international conference, TAMC 2007, Shanghai, China, May 22--25, 2007. Proceedings. Berlin: Springer (ISBN 978-3-540-72503-9/pbk). Lecture Notes in Computer Science 4484, 542-553 (2007). ZBL1198.03043.
Namely, they give a parametrized family of similar problems such that the collection of parameters for which Collatz-for-those-parameters is true, is $\Pi^0_2$-complete and hence not $\Pi^0_1$.
They can do this without thereby solving the Collatz problem, just like Matiyasevich et al. could show that solvability of diophantine equation was $\Sigma^0_1$-complete, without thereby solving any particular equation themselves.
If Collatz could somehow be simplified to a $\Pi^0_1$ form then quite plausibly the generalized version could too by the same argument (whatever that hypothetical argument would be) but that Kurtz and Simon show will not happen.
Best Answer
I am pretty sure that the answer to the question is no: no two of those big conjectures are known to imply the third. But I feel somewhat sheepish giving this as an answer: what evidence can I bring forth to support this, and if nothing, why should you believe me?
The only thing I can think of is that in the function field case, ABC and GRH are fully established, but only parts of BSD are known.
(Maybe I should also admit that I didn't know anything about the connection between ABC and bounds on Shafarevich-Tate groups of elliptic curves in terms of the conductor until I glanced just now at the paper of Goldfeld the OP linked to. The fact that you can build examples of large Sha from triples of integers with large ABC exponent is amazing to me.)
Addendum: I feel especially confident that ABC and GRH do not imply BSD, at least not the part of BSD that asserts finiteness of Shafarevich-Tate groups. The first two conjectures are essentially analytic in nature, whereas the finiteness of Sha is deeply arithmetic. It seems extremely unlikely.
Moreover, ABC is really hard, in the sense that for all of the results of the form "X implies ABC" that I've ever seen, X includes a statement which is ABC-like in the sense that it gives a uniform bound on one arithmetic quantity in terms of another. For example, ABC is known to be of a similar flavor to the Szpiro Conjecture (and implies it), but so far as I know it is only known to be implied by a more-explicitly-ABC-like Modified Szpiro Conjecture. Admittedly bounding Sha in terms of the conductor, as in Goldfeld's work, is only vaguely ABC-like, but to an arithmetic geometer like me these bounds still feel very "analytic"; I can't see any connection at all between this and BSD. So I doubt that GRH (let me say ERH, so that I more or less know what I'm talking about -- i.e., Dedekind zeta functions) plus BSD is known to imply ABC.