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.
Best Answer
the answers.com page you mentioned uses the following formula: $$L_{Planck} = \sqrt{\frac{Gh}{2\pi c^3}}$$ Note that there is the $2\pi$ factor in the denominator - so $h/2\pi$ may be simplified as the usual $\hbar$. They probably weren't able to type this character, or wanted to avoid terminology and symbols that are only known to physicists. But there is no numerical error on the answers.com page. At any rate, the definition above is equivalent to $$L_{Planck} = \sqrt{\frac{G\hbar}{c^3}}$$ which is the usual "unreduced" Planck length. See Wikipedia for the same formula:
http://en.wikipedia.org/wiki/Planck_length
Numerically, it's $1.6 \times 10^{-35}$ meters. (Update: The Oxford Dictionary of English has a wrong formula - they omitted $2\pi$ and forgot to cross the $h$, too. But they clearly mean the same Planck length.) Sometimes, people also use the "reduced" Planck length which is more fancy and "professional" in a sense: $$L_{Planck,reduced} = \sqrt{\frac{8\pi G\hbar}{c^3}}$$ Note that the $8\pi$ in the numerator may also be merged with the $\hbar$ to get $4h$ back - so the reduced Planck length is twice (because of the square root) the wrong Planck length that you would get by using $h$ instead of $\hbar$. But what is the real reason why $8\pi$ was added there?
The reason why $8\pi G$ appears instead of $G$ is because in some sense, $8\pi G$ is more natural a constant than $G$: this discussion is analogous to the treatment of $4\pi$ in electrodynamics. The constant $8\pi G$ is natural because the Einstein-Hilbert action is $$S_{EH} = \int d^D x \frac{1}{16\pi G} R\sqrt{-g} $$ The most natural coefficient would be $1/2$ instead of $1/16\pi G$ which makes it natural to set $8\pi G=1$. The reduced Planck length is somewhat longer (five times or so) - less extremely tiny. Even more often, particle physicists talk about the Planck energy and the reduced Planck energy which are close to $10^{19}$ and $10^{18}$ GeV, respectively.
The convention for the constant $G$ was originally chosen by Newton who wanted to write the gravitational force as $GMm/r^2$. Well, it would be more natural to have the factor of $4\pi$ or $8\pi$ in the denominator, $\Gamma Mm/8\pi r^2$. You can see that $\Gamma$ is simply $\Gamma=8\pi G$, and it would be natural to set $\Gamma$ equal to one.
I hope that I don't have to explain why $\hbar$ is more natural than $h$ for adult physicists. The "laymen" versions of the formulae may be simpler with $h$ - but they deal with wavelength etc. Adult physicists know that the wavelength of the sine is proportional to $2\pi$. And the most fundamental equations, such as the Schrödinger's equation or the commutators of $[x,p]$, take a simpler form in terms of $\hbar$ than $h$, of course.
Back to $G$: people had to choose the convention how to normalize $G$ in higher dimensions. The usual convention, as implicitly used above, is that the Einstein-Hilbert action always has the coefficient $1/16\pi G$. That implies that in $D$ spacetime dimensions, the force won't be $GMm/r^{D-2}$ but it will have some $D$-dependent numerical coefficients in it.
Best wishes Lubos