I hesitate a little to post this, since it probably not a full answer. But I feel than an important aspect is missing in the others and since comments are limited in size and not-editable I will post it like this.
It is true that the t-Test will generally give you good results if the sample size is large enough. What is large enough depends on the underlying distribution. Even so, it is possible to have a sample large enough for the CLT to kick in for the mean so that the nonparametric Wilcoxon test and the t-Test still give very different answers and are both right. That is because the t-Test tests the means and Wilcoxon test the medians. And in non-symmetric distributions these can differ.
In this case you don't just have to check whether it is valid to use these tests. As said, it is very possible that both will give reliable results. You also need to think about what you want to know. A typical example is income which can have a very high mean and a much larger median.
I say this because I do not know what you have a sample of, what interests you and what the deviations from normality look like. In many cases the distributions are symmetrical enough so that both tests answer more or less the same question. Sometimes they answer very different questions and this is not linked to the validity of the answers.
In your case I suspect that this is probably not the case, but it might help anyhow. As said, a difference between a p-value of 0.05 and 0.07 is not significant.
EDIT: I decided to expand on this even more, due to comments. It is true that we compare two samples. It is still the case that the Wilcoxon (even the Rank-Sum) test looks for a median shift > 0 and t.test looks for a mean shift > 0. Note that the median shift is not the shift of medians. Generate data in R like this:
x1 <- 100 + 0.01*rnorm(1000) #Effectively constant, with some jitter to avoid ties
shift.down <- seq(-10,0, by = 10/499)
shift.up <- seq(0,100, by = 100/499)
x2 <- 100 + c(shift.down, shift.up)
mean(x1-x2) # will be significant
wilcox.test(x1,x2) #will be insignificant
This works because both the median shift and the shift of medians is zero.
x1 <- rnorm(1000)
x2 <- rnorm(1000)
x2[x2>0] <- x2[x2>0]^4
will give a significant wilcoxon test since we have a median shift even though we do not have a shift of medians.