The main property of a logarithmic scale is that multiplication in the original scale gets converted into addition in the logarithmic scale. This property holds for decibels (dB) as well.
The example of the amplifier you give summarizes this pretty well. In the original scale, a signal in Watts $P$ is amplified by a dimensionless factor $A$, so the new signal $P'=AP$. Then in terms of the logarithmic quantities $L=\log P$, $a=\log A$, $L'=\log P$ (the base of the log doesn't really matter), we have $L' = a + L$.
Now, there is something a bit sneaky I slipped into the above paragraph. Namely, while $A$ is a dimensionless quantity so $a=\log A$ is well defined, $P$ and $P'$ have dimensions of Watts and so $\log P$ and $\log P'$ are not well defined. Therefore, for power, we need to introduce a reference scale $P_{\rm ref}$, and really $L=\log (P/P_{\rm ref})$ and $L'=\log (P'/P_{\rm ref})$. Since $A$ is dimensionless, it is ok to measure $a$ in ${\rm dB}$, but since $P$ and $P'$ are not dimensionless, we need to include the reference scale, and this choice of reference scale is reflected in the choice of unit, we use ${\rm dBm}$ instead of ${\rm dB}$ to indicate that we are using a reference power of $1\ {\rm mW}$.
Now we come to a key point. The whole idea of units to begin with is to let us keep track in changes in arbitrary reference scales. For example, if we convert from using a meter to a kilometer as a reference length, we know to divide all lengths by $1000$ to keep our equations consistent. Well, the key property of logs is that this rescaling will be converted to an addition. So changing the reference scale from say $1\ {\rm mW}$ to $1\ {W}$ will involve subtracting $\log 1000$ from all quantities with units of ${\rm dBm}$. In that sense, logarithmic units are an abuse of notation; they don't obey the same scaling rules as normal units. However, their transformation properties can be derived from the normal ones in a simple way.
It is perfectly fine mathematically to combine logarithmic quantities with different reference scales, so long as we keep track of all the reference scales involved. For example, we could consider Ohm's law
$$
V = IR
$$
taking a log of both sides and introducing reference scales,
$$
\log\left(\frac{V}{V_{\rm ref}}\right) = \log\left(\frac{I}{I_{\rm ref}}\right) + \log\left(\frac{R}{R_{\rm ref}}\right)
$$
where we should have $V_{\rm ref}=I_{\rm ref} R_{\rm ref}$ for consistency. If you defined units $\rm dBV$, $\rm dBA$, and $\rm dB\Omega$, then you could write expressions like
$$
15\ {\rm dBV} = 10\ {\rm dBA} + 5\ {\rm dB\Omega}
$$
and the logarithmic units would work out. If the base of the log was 10, then (up to factors of 20 or whatever) this would correspond to the equation
$$
10^{15} {\rm V} = 10^{10}\ {\rm A} \times 10^5{\rm \Omega}
$$
Now, I can't think of any practical situation where expressing Ohm's law in the logarithmic form would elucidate any physics, but in principle you could do it.
In practice, when one uses logarithmic scales, I think it is most common to multiply a dimensionful quantity by a dimensionless quantity, or two dimensionless quantities together. Therefore, you would naturally expect to see expressions like $10\ {\rm dBm} + 5\ {\rm dB} = 15\ {\rm dBm}$ or $10\ {\rm dB} + 5\ {\rm dB} = 15\ {\rm dB}$ "in the wild".
I believe I have now explained the principles behind your first two examples.
Your third example is of a somewhat different character. In fact ${\rm dBA}$ is not really a logarithmic unit, but a weighted logarithmic unit. Instead of taking the log of a natural physical quantity (the frequency), one is taking the $\log (R_A(f))$ for some complicated function $R_A(f)$ that is meant to model the human ear's frequency response. The introduction of this complicated $R_A(f)$ function means that the "algebra" associated the unit is complicated. I am not sure, but suspect that the "addition" symbol you defined in $60\ {\rm dBA} + 60\ {\rm dBA} = 63\ {\rm dBA}$ is defined so that the signal amplitudes add "normally" in terms of ${\rm dBm}$; if so, then to define the addition symbol acting on quantities with units ${\rm dBA}$, one needs to understand the properties of the complicated function $R_A(f)$. Thus, using ${\rm dBA}$ as a unit is in a sense an even worse abuse of notation than using ${\rm dB}$ as a unit, since one is using the notation to hide the complicated function $R_A(f)$. However, so long as one is clear on the conventions being used, there is no problem defining notation in this way.
Finally, you bring up the example of ${\rm dB/km}$. I actually believe this is a relatively easy extension of the idea of ${\rm dB}$. One just has to remember that under a change of scale of the signal amplitude, a ${\rm dB}$ changes by adding the log of the scale change, while under a change of scale of length, a ${\rm km}$ changes by multiplying the scale change.
