Product of a shifted Log-Normal and a Log-Normal distribution Let $X$ and $Y$ follow Log-Normal distributions, with $\ln X \sim \mathcal{N}(\mu_x, \sigma_x^2)$ and $\ln Y \sim \mathcal{N}(\mu_y, \sigma_y^2)$. $X$ and $Y$ are independent. Let $W = X (Y + c)$, where $c$ is a constant. Is $W$ still Log-Normally distributed? If not, can $W$ be approximated by a Log-Normal distribution?
I know that the product of Log-Normal distributions is Log-Normal. Unfortunately, $Y+c$ is no longer a Log-Normal, since its support is $[c, \infty)$. On the contrary, the support of $W$ is again $[0, \infty)$, which does not exclude the possibility that $W$ follows a Log-Normal distribution. 
Moreover, the numerical approximation of $W$ seems indeed to be Log-Normal - see the superposition of the histograms below, with $\mu_x = 2$, $\mu_y = 4$, $\sigma^2_x = \sigma^2_y = 1$, and $c = 5$. 

 A: Assuming the intent is that $X$ and $Y$ are independent, then $X(Y+c)$ is not lognormal.
(edit: consider that by Cramer's theorem $\log(W)$ will only be normal if both $\log(X)$ and $\log(Y+c)$ are, but the second term is not normal; Mathemagical mentions this argument below)
However, sums of bivariate lognormals are often very close to lognormal and $XY$ and $cX$ are jointly lognormal; for a fair range of values of the parameters, the approximation by a lognormal will probably be fairly decent.
With your particular example, a large simulation (say $10^6$ points) clearly shows that the lower tail of $\log(W)$ is a bit lighter and the upper tail a bit heavier than for a normal so you can tell without even worrying about the algebra. Repeated simulations show the same picture each time.

Nonetheless the lognormal approximation is excellent in that particular case, especially in the middle 98% or so of the distribution.
However, you can't always rely on it working without checking; it's not all that hard to come up with examples where the sum of lognormals is not close to lognormal. See some of the discussion and references in this post on CrossValidated -- The sum of independent lognormal random variables appears lognormal? for some additional information (though the question is about the independent case while $XY$ and $Xc$ are not independent, nonetheless the information is relevant)
Some papers for the correlated case:
Mehta, N.B., Wu, J.,  Molisch, A. F.,  and Zhang, J., (2007),
"Approximating a Sum of Random Variables with a Lognormal,"
IEEE Transactions on Wireless Communications, 6(7), 2690-2699. 
Rook CJ and Kerman MC, (2015),
"Approximating the Sum of Correlated Lognormals:  An Implementation"
arXiv:1508.07582 [q-fin.GN]
https://arxiv.org/ftp/arxiv/papers/1508/1508.07582.pdf
Lo, C.F. (2012),
The Sum and Difference of Two Lognormal Random Variables
Journal of Applied Mathematics Volume 2012, Article ID 838397, 13 pages
http://dx.doi.org/10.1155/2012/838397
A: Of course the numerical simulation cannot tell you that much since lot of distribtuions have a similar shape.
There are some trivial approaches for approximation for $c>0$:
$$
P(X(Y+c)\leq w)\leq \min\{P(XY\leq w),P(Xc\leq w\}\}.
$$
So the CDF is upper bounded by log-normal distributions. And for the lower bound for all $\delta\in(0,1)$:
$$
P(XY\leq w(1-\delta),cX\leq \delta w)\leq P(X(Y+c)\leq w),
$$
which has log-normal tail so no surprise that you see log-normal similarity in your numerical evaluations.

You can try to explicitly characterize the distribution, although not trivial:
$$
P(W\leq w)=P(X(Y+c)\leq w)=P(\ln X+\ln(Y+c)\leq \ln w)\\
=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-\dfrac{(x-\mu_x)^2}{2\sigma_x^2}}P(\ln(Y+c)\leq \ln w-x)\mathrm dx..
$$
Using the following
$$
P(Y+c\leq e^{w-x})=P(Y\leq e^{w-x}-c)=P(\ln Y\leq \ln(e^{\ln w-x}-c))=F_{\mu_y,\sigma^2_y}(\ln(e^{\ln w-x}-c)),
$$
we get
$$
P(W\leq w)
=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-\dfrac{(x-\mu_x)^2}{2\sigma_x^2}}F_{\mu_y,\sigma^2_y}(\ln(e^{\ln w-x}-c))\mathrm dx\\
=\int_{-\infty}^{\ln\frac wc}\frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-\dfrac{(x-\mu_x)^2}{2\sigma_x^2}}F_{\mu_y,\sigma^2_y}(\ln(e^{\ln w-x}-c))\mathrm dx.
$$
Even for the standard normal distribution, this will not be log-normal.
