Do there exist $0<b<a$ such that ${}_1\!F_1(a, b; -x)$ is nonnegative for $x >0$?

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The question I asked here is flawed because I didn't realize that in general the Kummer function ${}_1\!F_1(a, b; -x)$ can take negative values when $b<a$.

So now I'm wondering whether this is always the case, that is: for $a>0$ and $0<b<a$, does there always exist a value of $x$ such that ${}_1\!F_1(a, b; -x) < 0$ ? Or do there exist such values of $a$ and $b$ for which ${}_1\!F_1(a, b; -x)$ is nonnegative for all $x >0$ ?

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Since the Kummer equation is a Sturm–Liouville equation, the zeros of $F(A;B;z)$ are simple, so since $F$ is real for real $z$ if the parameters are real and $F(A;B;0)=1$, $F$ takes negative values if and only if it has a zero. So we need to determine when $F(a;b;-x)$ has at least one positive zero.

We have the relation $F(A;B;z) = e^z F(B-A;B;-z)$, and so the number of negative zeros of $F(A;B;x)$ is the same as the number of positive ones of $F(B-A;B;x)$.

Now, we are interested $0<b<a$, so we want the number of positive zeros of $F(b-a;b;x)$, i.e. the number of positive zeros of $F(A;B;x)$ with $A<0$ and $B \geq 0$. It is known that this is given by $\lceil -A \rceil $. (Noted in DLMF 13.9(i); simple proofs of these results are given by Skovgaard (1953), using that the derivatives of $F$ form a Sturm chain).

Therefore if $0<b<a$, the number of positive zeros of $F(a;b;-x)$ is $\lceil b-a \rceil $. Therefore, if $ \max{\{0,a-1\}} <b<a $, $F(a;b;-x)$ has no positive zeros, and so is always positive. There are plenty of other cases that could be considered, but this is the case you asked about; there's a diagram in Skovgaard's paper that summarises the general results.