The theorem says that;
If $G=(V(G),E(G))$ is connected graph on $n$-vertices where $n≥3]$ so that for $[[x,y∈V(G),$ where $x≠y$, and $deg(x)+deg(y)≥n$ for each pair of non-adjacent vertices $x$ and $y$ then $G$ is a Hamiltonian graph.
The simple meaning of the theorem is that, it says,
$$ \forall (\text{non-adjacent vertices pair } v,u ) (\operatorname{deg}(v) + \operatorname{deg} (w) ≥ n) \Rightarrow \text{Graph is Hamiltonian}$$
But this does not imply the reverse of it, that means,
$$ \text{Graph is Hamiltonian} \nRightarrow \\\forall (\text{non-adjacent vertices pair } v,u ) (\operatorname{deg}(v) + \operatorname{deg} (w) ≥ n) $$
No, there is no such example.
It is easier to ask: is there an $n$-vertex graph $G$ such that $\deg(u)+\deg(v) = n-1$ for one non-adjacent pair $u,v$, and $\deg(u)+\deg(v) \ge n$ for all other non-adjacent pairs, which does not have a Hamiltonian cycle? If the graph in your question exists, then by the adding-$w$ trick, this one must as well.
Suppose such a $G$ exists. Then $G+uv$ satisfies the hypotheses of Ore's theorem and has a Hamiltonian cycle. Either that cycle exists in $G$, or else $G$ at least has a $u-v$ Hamiltonian path: a path $(x_1, x_2, \dots, x_n)$ with $x_1 = u$ and $x_n = v$.
Suppose that $\deg(u) \ge 2$ (if not, look at $v$ instead of $u$), and let $x_i$ with $i>2$ be any neighbor of $u = x_1$. Then there is also a Hamiltonian path $(x_{i-1}, x_{i-2}, \dots, x_1, x_i, x_{i+1}, \dots, x_n)$ - and in this Hamiltonian path, $\deg(x_{i-1}) + \deg(x_n) \ge n$.
Now we can follow the standard proof of Ore's theorem to show that this Hamiltonian path can be turned into a Hamiltonian cycle. We conclude that $G$ is Hamiltonian; a single pair with $\deg(u)+\deg(v)=n-1$ is not enough to break the theorem.
In fact, a stronger argument can be made. Suppose that $G$ has fewer than $n-2$ non-adjacent pairs $u,v$ with $\deg(u)+\deg(v)=n-1$; for all other non-adjacent pairs $u,v$ the degree sum is $\deg(u)+\deg(v) \ge n$.
Then $G$ has a Hamiltonian path $(x_1, x_2, \dots, x_n)$ and by the argument above, we can turn this path into a total of $\deg(x_1) \cdot \deg(x_n)$ different Hamiltonian paths. Since $x_1, x_n$ cannot be adjacent, $\deg(x_1) + \deg(x_n) \ge n-1$, so $\deg(x_1) \cdot \deg(x_n) \ge n-2$. Therefore at least one of the Hamiltonian paths we get must end at vertices whose degree sum is at least $n$, and then the argument of Ore's theorem turns it into a Hamiltonian cycle.
The bound $n-2$ is tight: otherwise, take a $K_{n-1}$, and add a leaf vertex adjacent to one of its vertices. Here, there is no Hamiltonian cycle, and only $n-2$ pairs of non-adjacent vertices, each with degree sum $n-1$.
Similarly, for the Hamiltonian path question, we can take a $K_{n-1}$ together with an isolated vertex. Here, there is no Hamiltonian path, and only $n-1$ pairs of non-adjacent vertices, each with degree sum $n-2$. This is best possible.
Best Answer
The proof is by a form of induction, namely maximal counterexample. Let $S$ denote the set of all graphs on $n$ vertices, that violate the theorem. If $S$ is empty, we are happy as the theorem is true. If $S$ is nonempty, it must be finite, as there are only finitely many graphs on $n$ vertices. Hence, if we consider the ordering on $S$ based on number of edges, we can find a maximal example in $S$. (rest of theorem) Contradiction! Hence $S$ cannot be nonempty.