There is really just one underlying principle here: Any equation we write down should not depend on any arbitrary choices we made in order to define the quantities. All the examples you can discuss can be understood in this principle.
Can't add a vector and a scalar. Well, of course a vector is three numbers and a scalar is one number, so, for example, $\mathbf{v} + v$, , where $v$ is a speed, i.e. a scalar with units of velocity, doesn't even make mathematical sense. But we could add imagine adding one component of a vector to a scalar, i.e. $v_z + v$. But, this quantity shouldn't appear in a fundamental law of physics because our choice of what axis to call the $z$ axis is completely arbitrary, and if we made a different choice our equations would look different. But this is situation-dependent. For example, if we were discussing physics in a background uniform gravitational field, then we can use a convention where $z$ points along the direction of gravitational field. This is not arbitrary because the gravitational field sets a preferred direction. By declaring that we are going to call that particular direction the "z-direction'', it makes sense that any equations we then write down will only hold for that particular choice of $z$-axis. That is why the equation for the gravitational potential energy in a gravitational field, $U = mgz$, is valid even though $z$ is a component of the displacement vector $\textbf{r}$. However, you can still translate this equation into one that is valid for any choice of axes, namely $U = m \textbf{g} \cdot \textbf{r}$, where $\textbf{g}$ is the gravitational field vector.
Can't add numbers with different units. The point is that we normally work in units of physics which are chosen completely arbitrarily. If time $t$ is measured in seconds and position $x$ is measured in meters, then it
makes no sense to write down an equation involving $x+t$ because this equation would depend on our definition of "second" and "meter", and there is no reason why the laws of physics should depend on the second being defined to be 9,192,631,770 times the period of some radiation mode of a cesium atom. But, if we choose to work in natural units, then this is not an arbitrary choice because, as the name suggests, natural units are uniquely determined given fundamental constants of physics. In natural units, there is nothing wrong with writing an equation involving $x+t$, because we remember that we have made a special choice of units, and the equation will hold only in those units.
Of course, any equation that you can write in natural units can still be translated into arbitrary units. Take Einstein's famous mass-energy equivalence. In natural units ($c=1$) it states that $E = m$. Obviously, in arbitrary units, this is a bad equation because if $E$ is measured in Joules, and $m$ is measured in kg, then it would depend on the definitions of Joules and kg. But that's fine, because this equation only holds in natural units. Its translation into arbitrary units is $E = mc^2$, and the units now match up.
Can't add covariant 4-vectors and contravariant 4-vectors. Again, this is because in special relativity, in order to write down components of vectors we have to make an arbitrary choice of coordinate directions in space-time. Equations we write down in special relativity shouldn't depend on this choice, and this prevents us from adding covariant 4-vectors and contravariant 4-vectors because they transform differently when you change coordinate directions in space-time.
Can't add things in different vector spaces. This is just because, if $\mathbf{v}$ is in one vector space and $\mathbf{w}$ is in a totally different vector space, then you wrote an equation involving $\mathbf{v} + \mathbf{w}$ then it would depend on how you relate the bases between the two vector spaces, which -- given that they are totally different spaces -- there is no way to do non-arbitrarily.
Best Answer
The short answer is that it can, if $M = 1 = M^{-1}$. In this way of looking at it, all quantities in Planck units are pure numbers.
The longer answer is that there are two different ways of thinking about natural unit systems.
Natural unit systems in terms of standard units
One of them, and perhaps the easier one to understand, is that you're still working in a "traditional" unit system in which distinct units for all quantities exist, but the units are chosen such that the numerical values of certain constants are equal to 1. For example, if you want to set $c = 1$, you're not literally setting $c = 1$, you're actually setting $c = 1\,\frac{\text{length unit}}{\text{time unit}}$. Length and time don't actually have the same units in this interpretation; they're equivalent up to a multiplication by factors of $c$. In other words, it's understood that to convert from, say, a time unit to a length unit you multiply by $c$, and so that is left implicit.
In order to do this, of course, you have to choose a length unit and time unit which are compatible with this equation. So you couldn't use meters as your length unit and seconds as your time unit, but you could use light-seconds and seconds, respectively.
If you want to set multiple constants to have numerical values of 1, that constrains your possible choices of units even further. For example, suppose you're setting $c$ and $G$ to have numerical values of 1. That means your units have to satisfy both the constraints
$$\begin{align} c &= \frac{\text{length unit}}{\text{time unit}} = \frac{\ell_G}{t_G} & G &= \frac{(\text{length unit})^3}{(\text{mass unit})(\text{time unit})^2} = \frac{\ell_G^3}{m_Gt_G^2} \end{align}$$
where I've introduced $\ell_G$, $t_G$, and $m_G$ to stand for the length, time, and mass units in this system, respectively. You can then invert these equations to solve for $\ell_G$, $t_G$, and $m_G$ in terms of $c$ and $G$ - but as you can probably tell, the system of equations is underdetermined. It still gives you the freedom to choose one unit to be part of your unit system, such as
$$\text{kilogram} = \text{mass unit} = m_G$$
Having made that choice, you can now solve for $m_G$, $\ell_G$, and $t_G$ in terms of $c$, $G$, and $\text{kilogram}$ (or whatever other choice you might have made; each choice gives you a different unit system).
Running through the math for this gets you
$$\begin{align} m_G &= 1\text{ kg} & \ell_G &= \frac{G (1\text{ kg})}{c^2} & t_G &= \frac{\ell_G}{c} = \frac{G (1\text{ kg})}{c^3} \end{align}$$
Now you can plug in values of $G$ and $c$ in, say, SI units, and get conversions from SI (or whatever) to this unit system. Note that, as I said, length does not literally have the same units as time or mass, but you can convert between the length unit, time unit, and mass unit by multiplying by factors of $G$ and $c$, constants which have numerical values of 1. In a sense, you can consider this multiplication by $G^ic^j$ as analogous to a gauge transformation, i.e. a transformation that has no effect on the numerical value of a quantity, and the units of length, time, and mass are mapped on to each other by this transformation just as gauge-equivalent states are mapped on to each other by a gauge transformation in QFT. So it's more proper to say $L \sim T \sim M$; the dimensions are not equal, just equivalent under some transformation.
If you do the same thing but setting $c = \hbar = 1$ instead, remember what you're really doing is specifying that your units must satisfy the constraints
$$\begin{align} c &= \frac{\text{length unit}}{\text{time unit}} = \frac{\ell_Q}{t_Q} & \hbar &= \frac{(\text{length unit})^2(\text{mass unit})}{(\text{time unit})} = \frac{\ell_Q^2m_Q}{t_Q} \end{align}$$
($Q$ is for "quantum" because these are typical QFT units), and then running through the math, again with $m_Q = 1\text{ kg}$, you get
$$\begin{align} m_Q &= 1\text{ kg} & \ell_Q &= \frac{\hbar}{(1\text{ kg})c} & t_Q &= \frac{\ell_Q}{c} = \frac{\hbar}{(1\text{ kg})c^2} \end{align}$$
Again, the units are not literally identical, but $\ell_Q \sim t_Q \sim m_Q^{-1}$ under multiplication by factors of $\hbar$ and $c$.
Of course, your third constraint doesn't have to be a choice of one of the fundamental units. You can also choose a third physical constant to have a numerical value of 1. To obtain Planck units, for example, you would specify
$$\begin{align} c &= \frac{\text{length unit}}{\text{time unit}} = \frac{\ell_P}{t_P} \\ \hbar &= \frac{(\text{length unit})^2(\text{mass unit})}{(\text{time unit})} = \frac{\ell_P^2m_P}{t_P} \\ G &= \frac{(\text{length unit})^3}{(\text{mass unit})(\text{time unit})^2} = \frac{\ell_P^3}{m_Pt_P^2} \end{align}$$
You can tell that this is no longer an underdetermined system of equations. Solving it gives you
$$\begin{align} m_P &= \sqrt{\frac{\hbar c}{G}} & \ell_P &= \sqrt{\frac{\hbar G}{c^3}} & t_P &= \sqrt{\frac{\hbar G}{c^5}} \end{align}$$
Here, since you've set three constants to have numerical values of 1, your three fundamental Planck units will be equivalent up to multiplications by factors of those three constants, $G$, $\hbar$, and $c$. In other words, multiplication by any factor of the form $G^i\hbar^jc^k$ is the equivalent to the gauge transformation I mentioned earlier. You can tell that all these units are equivalent under such a transformation, but more than that, all powers of them are equivalent! In particular, you can convert between $M$ and $M^{-1}$ by multiplying by constants whose numerical value in this unit system is equal to 1, and thus it's not a problem that $M \sim M^{-1}$ here.
Unit systems as vector spaces
Another way of understanding unit systems, which is kind of a logical extension of the previous section, is to think of them as a vector space. Elements of this vector space correspond to dimensions of quantities, and the basis vectors can be chosen to correspond to the fundamental dimensions $L$, $T$, and $M$. (Of course you could just as well choose another basis, but this one suits my purposes.) You might represent
$$\begin{align} L &\leftrightarrow (1,0,0) & T &\leftrightarrow (0,1,0) & M &\leftrightarrow (0,0,1) \end{align}$$
Addition of vectors corresponds to multiplication of the corresponding dimensions. Derived dimensions correspond to other vectors, like
$$\begin{align} [c] = LT^{-1} &\leftrightarrow (1,-1,0) \\ [G] = L^3M^{-1}T^{-2} &\leftrightarrow (3,-2,-1) \\ [\hbar] = L^2MT^{-1} &\leftrightarrow (2,-1,1) \end{align}$$
In this view, setting a constant to have a numerical value of 1 corresponds to projecting the vector space onto a subspace orthogonal to the vector corresponding to that constant. For example, if you want to set $c = 1$, you project the 3D vector space on to the 2D space orthogonal to $(1,-1,0)$. Any two vectors in the original space which differ by a multiple of $(1,-1,0)$ correspond to the same point in the subspace - just like how, in the previous section, any two dimensions which could be converted into each other by multiplying by factors of $c$ could be considered equivalent. But in this view, you can actually think of the two dimensions as becoming the same, so that e.g. length and time are actually measured in the same unit.
Since in Planck units you set three constants to have a numerical value of one, in the dimensions-as-vector-space picture, you need to perform three projections to get to Planck units. Performing three projections on a 3D vector space leaves you with a 0D vector space - the entire space has been reduced to just a point. All the units are mapped to that one point, and are the same. So again, $M$ and $M^{-1}$ are identical, and there's no conflict.