After further thought, I realized that most of these congruences are all immediate consequences of the well-known property of Fermat quotients

$$ q_p (a^n) \equiv n q_p (a) \pmod{p} $$

along with the following property: if $p$ is an odd prime and $r$ is an integer and we write $r p = a \pm b$ with $(p,ab)=1$, then

$$ a q_p (b) \pm b q_p(a) \equiv r \pmod{p} $$

Together, these show that if $rp = a^m \pm b^n$, then

$$ a^m nq_p (b) \pm b^n m q_p (a) \equiv r \pmod{p} $$

What shocks me is that these latter congruences have a short completely elementary proof but are not mentioned in a lot of the literature on Fermat quotients. I'd add them to the basic properties of Fermat quotients on Wikipedia, except that I doubt one post on a forum constitutes a reliable source.

To prove the second congruence above, note that

$$ b q_p(a) p = b (a^{p-1} - 1) = b \left( (rp \mp b)^{p-1} - 1 \right) \equiv b \left( \mp b^{p-2} r (p-1) +q_p(b) \right) p \pmod{p^2} $$

and thus

$$ b q_p(a) \equiv \pm b^{p-1} r + b q_p(b) \equiv \pm r + b q_p(b) \pmod{p} $$

Since $a \equiv \mp b \pmod{p}$ it follows that

$$ \pm b q_p(a) \equiv r - a q_p(b) \pmod{p}$$

from which the second congruence above follows.

Edit: I finally found a reference! I knew this simple of a result could not be new! The earliest I can find is from a journal called "L'Intermediaire des Mathematiciens", of which I had not previously heard. It seems fascinatingly to be a publication where people could pose short questions hoping that others would answer --- perhaps like a paper version of MO! Does anyone know the history of this journal? Was its function indeed like an early version of MO?

On p.121, volume 20, published 1913, one E. Dubouis poses the question (numbered 3764): How does one prove the theorem of Mirimanoff (C.R. 24 January 1910): `If p is a prime number of the form $2^a 3^b \pm 1$ or of the form $\pm 2^a \pm 3^b$, then at least one of the numbers $2^{p-1} - 1$ and $3^{p-1} -1$ is not divisible by $p^2$?' "

Later in that same volume, p.206, one finds an answer by Endmund Landau: "One can easily prove the more general fact that the following three lines lead to a contradiction:

$$ mp = x+y $$
$$ p \text{ odd } > 1 \text{ and not dividing } m $$
$$ x^{p-1} \equiv y^{p-1} \equiv 1 \pmod{p^2} $$

(These congruences are satisfied for $x=2^a 3^b$, $y = \pm 1$ and for $x = \pm 2^a$, $y = \pm 3^b$, if $2^{p-1} \equiv 3^{p-1} \equiv 1 \pmod{p^2}$; taking $m=1$ and $p$ prime, we recover the question posed by M.[sic] Dubouis.) We have:

$$ 1 \equiv x^{p-1} \equiv (mp-y)^{p-1} \equiv -(p-1) mpy^{p-2}+y^{p-1} \equiv -(p-1) mpy^{p-2} + 1 \pmod{p^2}$$

and $p$ divides $(p-1) m y^{p-2}$, hence $m$, contrary to hypothesis."

So although E. Landau doesn't explicitly state the second congruence above, the proof I gave is almost the same as Landau's.