Why is the equilibrium constant unitless?
There are a few separate issues here to keep in mind: 


$K_c$ (the equilibrium constant in terms of concentrations) is defined as 
$$K_c = \prod {c_i}^{\nu_i} \tag{1}$$
Agreement between $K_c$ and the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional rate law, and at equilibrium. 
When dealing with gases, $K_c$ cannot generally be equated with $K_p$ (the equilibrium constant in terms of partial pressures). They may be numerically equal under some circumstances. For instance, they are numerically equal in reactions involving only ideal gases, provided the number of gaseous molecular entities remains constant during the reaction.
The (standard) equilibrium constant $K$, which simplifies to $K_a$ (under standard fugacity and some other conditions, agreements) which is defined as $$K_a = \prod a_i^{\nu_i}, \tag{2a}$$ cannot generally be equated with $K_c$. They may be numerically equal under some circumstances. $K$ itself is actually defined as follows:  $$\Delta_\pu{r} G^\circ=-RT\log(K) \tag{3}$$ whereas, strictly speaking, $K_c$ can't be used in this relation (but see below). The equilibrium constant $K$ cannot have units or this equation will not make sense.  In terms of concentrations,$$K_a = \prod \left(\frac{\gamma_ic_i}{c_i^\circ}\right)^{\nu_i} \tag{2b}$$ (To be nitpicky, one should use molality-based activities instead of amount concentration-based $c$ when saying that $K$ reduces to $K_a$).  When activity coefficients are all unity (ideal conditions), it might be that $$K_a \approx \frac{K_c}{K_c^\circ}= \prod \left(\frac{c_i}{c_i^\circ}\right)^{\nu_i} \tag{2c}$$ Strictly speaking, the equilibrium constant $K$ should remain unchanged with a change in the units used to describe concentrations. Units have no place in the constant because they cancel when ratios to reference standard states are computed (see Eq. (2)). $K_c$ and $K$ differ by the product  $$K_c^\circ = \prod ({c_i^\circ})^{\nu_i}\tag{2d}$$ (ignoring, once again, activity coefficients; standard fugacity). This difference is the source of the dangling units leading to the question in the OP.


You should therefore be careful equating $K_c$ with the ratio $k_{\pu{fwd}}/k_{\pu{rev}}$. And, assuming they are equal, the concentration equilibrium constant cannot generally be equated with $K$. However you can divide $K_c$ by $K_c^\circ$, and, assuming activity coefficients are unity, the result might be the thermodynamically correct unitless equilibrium constant (if standard fugacity), equivalent to that which appears in usual thermodynamic derivations. 

Sloppy or not, associating units with an equilibrium constant is common practice. Attaching units to the equilibrium constant during reporting is a useful mnemonic when performing some computations, but it should be kept in mind that such a constant may not fulfill the requirements for computation of an associated standard free energy change, unless first modified by a ratio product of the standard concentrations. 

Note that some of this is also explained in an answer to this post. IUPAC likes to use the convention standard equilibrium constant when referring to the thermodynamic equilibrium constant.