I believe that the weaker question can be proved by induction on $n$. The case $n=1$ is clear. Now assume for $n-1$ and expand the $n\times n$ determinant by the first row. At least one of the terms in the expansion must be odd. Thus the original matrix $A$ has an $(n-1)\times (n-1)$ submatrix $B$, say consisting of entries not in row 1 or column $j$, with odd determinant such that $A_{1j}=1$ By induction we can change some of the 1's in $B$ to odd integers so that the new matrix $B'$ satisfies det$(B')=1$. Let $A'$ be $A$ after replacing $B$ with $B'$. Now det$(A')= A_{1j} +$ terms not involving $A_{1j}$, say det$(A')=A_{1j}+c$. Since $A_{1j}=1$ and det$(A')$ is odd, it follows that $c$ is even. Hence we can replace $A_{1j}$ with the odd integer $1-c$ so that the resulting matrix has determinant 1.

I believe that the weaker question can be proved by induction on $n$. The case $n=1$ is clear. Now assume for $n-1$ and expand the $n\times n$ determinant by the first row. At least one of the terms in the expansion must be odd. Thus the original matrix $A$ has an $(n-1)\times (n-1)$ submatrix $B$, say consisting of entries not in row 1 or column $j$, with odd determinant such that $A_{1j}=1$. By induction we can change some of the 1's in $B$ to odd integers so that the new matrix $B'$ satisfies det$(B')=1$. Let $A'$ be $A$ after replacing $B$ with $B'$. Now det$(A')= A_{1j} +$ terms not involving $A_{1j}$, say det$(A')=A_{1j}+c$. Since $A_{1j}=1$ and det$(A')$ is odd, it follows that $c$ is even. Hence we can replace $A_{1j}$ with the odd integer $1-c$ so that the resulting matrix has determinant 1.