No, they do not have the same definition, though you may have been given the definitions in forms that are superficially very similar. One common definition of the $T_1$ property is that $X$ is $T_1$ if for each pair of distinct points $x$ and $y$ there is an open set $U$ such that $x\in U$ and $y\notin U$. What you may not realize right away is that because this applies to every pair of points, it applies to $y$ and $x$ as well, so that there is also an open set $V$ such that $y\in V$ and $x\notin V$. In other words, given two points in a $T_1$-space, each has an open neighborhood that does not contain the other. This is equivalent to saying that all singletons in $X$ are closed, but this does require a proof.
A space $X$ is $T_0$ if for each pair of distinct points $x$ and $y$ in $X$ at least one of $x$ and $y$ has an open neighborhood that does not contain the other; it doesn’t have to work both ways. Thus, the topology
$\Big\{\varnothing,\{a\},\{a,b\},\{a,b,c\}\Big\}\tag{1}$
is a $T_0$ topology on $\{a,b,c\}$: given any two points in the space, the one that comes earlier in the alphabet has an open neighborhood that does not contain the other. It is not a $T_1$-space, however, because (for instance) $c$ has no open neighborhood that does not contain $a$: the only open neighborhood of $c$ is $\{a,b,c\}$. Thus, when we consider the pair of points $c$ and $a$ (in that order), there is no open $U$ such that $c\in U$ and $a\notin U$. And you can also see that the only closed singleton in this space is $\{c\}$: the sets $\{a\}$ and $\{b\}$ are not closed in this topology.
The only $T_1$ topology on $\{a,b,c\}$ is the discrete topology, in which every subset of $\{a,b,c\}$ is open.
Added: It’s only mildly painful to find all of the $T_0$-topologies on $X=\{a,b,c\}$. One is the discrete topology, but it’s also $T_1$ (and more). There are altogether $6$ $T_0$ topologies homeomorphic to $(1)$, one for each permutation of $X$. There are $6$ more $T_0$ topologies homeomorphic to
$\Big\{\varnothing,\{a\},\{a,b\},\{c\},X\Big\}\;;$
there are $3$ ways to choose the non-isolated point, and then $2$ ways to choose the isolated point with which to pair it. And that’s it: one $T_1$ topology, and $12$ topologies that are $T_0$ but not $T_1$, $6$ in each of two homeomorphism classes. All other topologies on $X$ fail to be even $T_0$.