Yes, if the invertible elements in $B(X)/K(X)$ are dense then the Fredholm operators are dense.
Let $x \in B(X)$ and consider $\pi(x) \in B(X)/K(X)$.
By density of the invertible elements $I$ in $B(X)/K(X)$, there exists a (Cauchy) sequence $y_n \in I$ such that $y_n \rightarrow \pi(x)$ as $n \rightarrow \infty$.
Using an argument similar to the one that shows that the quotient of a Banach space by a closed subspace is a Banach, we can obtain a Cauchy sequence $x_n \in B(X)$ such that $\pi(x_n) = y_n$.
Note that the $x_n \in B(X)$ are Fredholm since $\pi(x_n) \in I$.
Completeness of $B(X)$ then implies that $x_n$ converges to some $x_\infty \in B(X)$ which must satisfy $\pi(x_\infty) = \pi(x)$.
Thus, there exists some Fredholm operator $k \in K(X)$ such that $x_\infty + k =x$.
It follows that the sequence $x_n + k$ is a sequence of Fredholm operators that converges to $x$.
Note that this proof doesn't use anything special about Fredholm operators but rather follows simply from properties of Banach spaces and their quotients.
Also, the converse statement is straightforward to show.