The fact that
$$\int_E |f_n - f| \to 0 \Rightarrow \int_E |f_n| \to \int_E |f|$$
is actually a simple consequence of the fact that $f \mapsto \int_E |f|$ is a norm; in other words, it holds independently of the assumptions given. More specifically, in any normed space, the following inequality holds
$$ |\| x \| - \| y \| | \leq \| x - y \|.$$ The proof is easy: if $\| x \| \geq \| y \|$, write $\| x \| = \| x - y + y \| \leq \| x - y \| + \| y \|$, so that $0 \leq \|x\| - \| y \| \leq \| x - y \|$. If $\| y \| > \| x \|$, reverse the roles of $x$ and $y$, to get that $\| y \| - \| x \| \leq \| x -y \|$. Taking these inequalities together proves the result.
Hence $\| f_n - f\|_1 \geq |\| f_n\|_1 - \| f \|_1|$, and since the left hand side tends to $0$, so must the right hand side, since it is $\geq 0$. This proves the first implication.
For the converse, assume that $$ \int_E |f_n | \to \int_E |f|$$ and that $f_n \to f$ a.e. on $E$. We can, as Elan B. mentions, take $g_n = |f_n| + |f|$ and conclude using the triangle inequality that $|f_n - f| \leq g_n$. By assumption, we have $|f_n| \to |f|$ a.e. on $E$ (since, again, $|f_n - f | \to 0$ and $|f_n - f| \geq | |f_n| - |f| |$). Furthermore, we have $$ \int_E g_n = \int_E (|f_n| + |f|) \to \int_E 2|f|.$$ From the generalized DCT, we conclude that $$\int_E |f_n - f| \to 0,$$ as claimed. This proves (1).
If all that you assume is that $(f_n)_{n\in\Bbb N}$ converges pointwise to $f$, then all you know is that $\lim_{n\to\infty}f_n(x)=f(x)$ for each individual $x\in E$. But you cannot deduce from this that, given $\varepsilon>0$, you have $\bigl|f(x)-f_n(x)\bigr|$ for every sufficiently large $N$ and for all $x\in E$.
For instance, take $E=[0,1]$. For each $x\in E$ and each $n\in\Bbb N$, defined$$f_n(x)=\begin{cases}n^2x&\text{ if }x<\frac1{2n}\\-n^2\left(x-\frac1n\right)&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ if }x>\frac1n,\end{cases}$$Then $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function, but, for each $n\in\Bbb N$, $\int_0^1f_n(x)\,\mathrm dx=\frac14$.
Best Answer
Let $E=[0,1]$ and define $f_n(x)=(n+1)x^n$. Then $f=\lim f_n\equiv0$ a.e., while $\int_E f_n =1$.