Coulomb potential in 2D
The length scale $L$ has to be present in the denominator for dimensional reasons – only logarithms of dimensionless quantities are really "well-defined" unless one wants to introduce bizarre units such as the "logarithm of a meter".
On the other hand, the dependence on $L$ is largely trivial and unphysical for most purposes. Replace $L$ by $K$ and you will get $$ V_K(x,y) = -\ln \left( \frac{|\vec x|}{K} \right) = V_L(x,y) +\ln(K/L) $$ which only differs by the additive shift, $\ln(K/L)$ from the original potential you mentioned. Shifts of potentials by a constant are largely inconsequential. In particular, the gradient of $V$, $\nabla V$, isn't changed at all. To derive the simple claim about the shift above, I only used $\ln(A/B) = \ln(A)-\ln(B)$ a few times.
The Fourier transform of your potential may be derived by realizing what the Laplacian of the potential is. The Laplacian is the two-dimensional delta-function. In the momentum basis, it's equivalent to the identity $$ (p_x^2+p_y^2) \tilde V(p_x,p_x) = 1 $$ which is easily solved by $\tilde V =1/(p_x^2+p_y^2)$. However, the behavior of $\tilde V$ isn't quite well-defined for the point $p_x=p_y=0$ where one can add a multiple of a delta-function. This is because $$(p_x^2+p_y^2) \delta(p_x) \delta (p_y) = 0$$ so $\tilde V \to \tilde V + K \delta(p_x)\delta(p_y)$ transforms a solution into another solution. Of course, the two-dimensional delta-function in the momentum space is nothing else than the Fourier transform of the constant term $\ln(K/L)$ we discussed in the position basis so the two ambiguities are the same.
Now, you could think that the momentum-basis form of the potential, $1/(p_x^2+p_y^2)$, is unique because it has no length scale in it and no delta-function in it while we don't see a corresponding unique form of the position-basis potential – because the expressions with any length scale are equally good. But this is really an illusion. As a distribution, $1/(p_x^2+p_y^2)$ is ill-defined (in the very same sense as $\ln(x^2+y^2)$ would be in the position basis) and we must specify what its behavior near the origin is.
This ambiguity is the two-dimensional generalization of the subtleties connected with the one-dimensional "principal value" of $1/x$ as a distribution. $1/x$ multiplied by a test function is well-defined if we agree that the symmetric region $x\in(-\epsilon,+\epsilon)$ is removed. That's what we mean by the principal value.
On the other hand, if you compute the two-dimensional integral of $1/(p_x^2+p_y^2) f(p_x,p_y)$ where $f$ is continuous near the origin, you may switch to the polar coordinates where $r$ in $r\,dr\,d\phi$ is beaten by $1/r^2=1/(p_x^2+p_y^2)$ so you still have a divergent integral that has to be regulated. A way to regulate it is to cut if off and remove the disk $r<p_{\rm min}$ for some small $p_{\rm min}$. Such a cutoff induces an additive shift dependence that is logarithmic in the cutoff. For the same dimensional reasons as before, one has to take the logarithmic dimensionless so what we need to subtract (or add?) to erase most of the dependence on the cutoff is something like $$ f(0) \ln(p_{\rm min} / L_P) $$ where $L_P$ is the counterpart of $L$, the length scale you started with. Sorry if I omitted some dimensionless coefficients. Clearly, the change of $L_P$ is equivalent to redefining the distribution by an additive shift by $\delta^{(2)}(\vec p)\times L_P$ and $L_P \sim 1/L$ plays the same role of the scale we had before, in the position basis.