If nonadjacent $u,v$ in an $n$-vertex graph satisfy $\deg(u)+\deg(v)\ge n$, then adding $uv$ preserves Hamiltonicity in both directions

Every Hamilton cycle of $G$ is also a Hamilton cycle of $G+uv$, so Hamiltonicity of $G$ implies Hamiltonicity of $G+uv$.

Conversely, take a Hamilton cycle of $G+uv$. If it avoids the new edge $uv$, it is already a Hamilton cycle of $G$. If it uses $uv$, deleting that edge leaves a Hamilton path $v_1=u,v_2,\ldots,v_n=v$ in $G$.

For $1\le i\le n-1$, put $i$ in $A$ when $uv_{i+1}$ is an edge of $G$, and put $i$ in $B$ when $v_iv$ is an edge. Since $u$ and $v$ are nonadjacent, $|A|=\deg_G(u)$ and $|B|=\deg_G(v)$. Both lie in an $(n-1)$-element index set, while $|A|+|B|\ge n$, so $A\cap B\ne\varnothing$.

Choose $i\in A\cap B$. Then $v_1,v_2,\ldots,v_i,v_n,v_{n-1},\ldots,v_{i+1},v_1$ is a Hamilton cycle in $G$: the two joining edges are $v_iv$ and $uv_{i+1}$, and all other edges come from the path in step 1.2.

The cases in step 1.2 and the construction in step 2.1 prove that Hamiltonicity of $G+uv$ implies Hamiltonicity of $G$; step 1.1 proves the other direction.