= Solution
\b[The printed assertion for every cohomological degree is false.] The valid conclusion supplied by the stated <Hartogs extension theorem> is the degree-zero isomorphism. Indeed, around each removed point choose a small coordinate ball on which the <holomorphic vector bundle> $E$ is trivial. A <holomorphic section> on the punctured ball has finitely many holomorphic coefficient functions, each of which extends by <Hartogs extension theorem> when the complex dimension is at least two. Uniqueness of holomorphic extension makes these local extensions agree with the original section on overlaps, giving
$$
\boxed{H^0(X,\mathcal O(E))\cong H^0(X_0,\mathcal O(E|_{X_0})).}
$$
This also identifies the <direct image sheaf> $j_*\mathcal O(E|_{X_0})$ with $\mathcal O(E)$ for the inclusion $j:X_0\hookrightarrow X$. It does not identify higher <sheaf cohomology>.
Here is an explicit higher-degree obstruction. Take $X=\mathbb P^2$, remove $[0:0:1]$, and take the trivial line bundle. Cover $X_0$ by $U=\{x\ne0\}$ and $V=\{y\ne0\}$. On $U$ use coordinates $t=y/x$, $s=z/x$; on $V$ use $u=x/y=1/t$, $v=z/y=s/t$. Both charts are $\mathbb C^2$, and the overlap is $\mathbb C^*\times\mathbb C$. For every integer $m\geq2$, the <holomorphic function>
$$
g_m=\frac{z^m}{x^{m-1}y}=\frac{s^m}{t}
$$
defines a <Čech cocycle>. It cannot be a <Čech coboundary> $b(1/t,s/t)-a(t,s)$ with $a,b$ entire on their charts. Taking the coefficient of $s^m$, the term from $a$ has only nonnegative powers of $t$, whereas the term from $b$ has the form $t^{-m}b_m(1/t)$ and only powers at most $-m$. Neither can provide the coefficient $t^{-1}$. Uniqueness of <Laurent series> proves the contradiction. The same argument applied to each fibre degree proves that all the classes $[g_m]$, $m\geq2$, are linearly independent.
These classes remain nonzero in <Čech cohomology> of $X_0$, rather than merely of this cover. The low-degree <Mayer-Vietoris sequence for sheaf cohomology> injects the quotient of overlap sections by the two chart-section groups into $H^1(X_0,\mathcal O)$. Concretely, a <partition of unity> gives smooth $f_U,f_V$ with $f_V-f_U=g$; their common <Dolbeault operator> defines a global $(0,1)$-form. If that form were $\bar\partial$-exact, subtracting its global smooth primitive would make $f_U,f_V$ holomorphic and split $g$. The nonsplitting just proved therefore gives the same obstruction in <Dolbeault cohomology>, and the <Dolbeault theorem> identifies it with <sheaf cohomology>. Thus $H^1(X_0,\mathcal O)$ is infinite dimensional. In contrast, $H^1(\mathbb P^2,\mathcal O)$ is finite dimensional by compact <Dolbeault Hodge decomposition> for the <Fubini-Study form>. Hence \b[the two degree-one groups cannot be isomorphic], disproving the printed all-degree request already in complex dimension two.
In complex dimension one, even the degree-zero conclusion fails. Take $X=\mathbb P^1$, remove its point at infinity and use the trivial line bundle. Holomorphic functions on compact connected $\mathbb P^1$ are constant by the <maximum modulus principle>, whereas $X_0\cong\mathbb C$ has nonconstant entire functions such as $z$. Therefore \b[the restriction map is not onto even in degree zero].
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