$\newcommand\ep{\epsilon}$We have 1)$\implies$2) but 2)$\kern5pt\not\kern-5pt\implies$1). 

Indeed, for each real $a>0$, consider the bounded Lipschitz functions $f_a$ and $g_a$ defined by  
\begin{equation*}
	f_a(x):=a\wedge|x|,\quad g_a(x):=(-a)\vee(a\wedge x)
\end{equation*}
for real $x$, where $u\vee v:=\max(u,v)$ and $u\wedge v:=\min(u,v)$.

Suppose now that 1) holds. Take any real $a>0$ such that $P(|X|\le a/2)>0$. Note that 
\begin{multline*}
	P(|X|\le a/2,|X_n|>a)\le P(f_a(X)\le a/2,f_a(X_n)\ge a) \\ 
	\le P(|f_a(X_n)-f_a(X)|\ge a/2) \le E|f_a(X_n)-f_a(X)|/(a/2), 
\end{multline*}
by Markov's inequality. So, in view of 1), 
\begin{equation*}
	\sum_n P(|X|\le a/2,|X_n|>a)<\infty. 
\end{equation*}
Therefore and because of the condition $P(|X|\le a/2)>0$, 
\begin{multline*}
	\sum_n P(|X|\le a/2,|X_n|\le a)=\sum_n [P(|X|\le a/2)-P(|X|\le a/2,|X_n|>a)] \\ 
	=\sum_n P(|X|\le a/2)-\sum_n  P(|X|\le a/2,|X_n|>a)=\infty. 
\end{multline*}
Hence, 
\begin{equation*}
	\sum_n P(|X|\vee|X_n|\le a)=\infty. \tag{*}
\end{equation*}

Next, for any real $\ep>0$ 
\begin{multline*}
	\sum_n P(|X_n-X|>\ep,|X|\vee|X_n|\le a)
	\le\sum_n P(|g_a(X_n)-g_a(X)|>\ep) \\ 
	\le\sum_n E|g_a(X_n)-g_a(X)|/\ep<\infty,
\end{multline*}
by Markov's inequality and 1). 

So, in view of (*), 
\begin{multline*}
	\sum_n P(|X_n-X|\le\ep)\ge\sum_n P(|X_n-X|\le\ep,|X|\vee|X_n|\le a) \\ 
=\sum_n P(|X|\vee|X_n|\le a) 
-\sum_n P(|X_n-X|>\ep,|X|\vee|X_n|\le a) 
=\infty,
\end{multline*}
so that 2) holds. 
Thus, 1)$\implies$2). 

Now, as suggested in the comment by Martin Hairer, suppose that $X=0$ and $P(X_n=0)=P(X_n=1)=1/2$ for all $n$. Then $P(|X_n-X|\le\ep)\ge1/2$ for all $n$ and hence 2) holds. On the other hand, $E|f_1(X_n)-f_1(X)|=1/2$ and hence 1) does not hold (for $f(x)\equiv x$). Thus, 2)$\kern5pt\not\kern-5pt\implies$1).