Assuming the assertions your claim are true, it follows that $\mathrm{GL}_n(\mathbf{Z})$ is generated by 2 elements for all $n$; for $n$ odd (your missing case), the matrices $-s_1,s_3$ indeed generate $\mathrm{GL}_n(\mathbf{Z})$.

This follows from the following three facts:

1. Let $p$ be prime and $C_p=\langle c\rangle$ the cyclic group of order $p$. If $G$ is an arbitrary group with $\mathrm{Hom}(G,C_p)=0$, and if $G$ is generated by $k\ge 1$ elements, then so is $G\times C_p$. Proof: if $g_1,\dots,g_k$ generate $G$, then $(g_1,c),(g_2,1)\dots,(g_k,1)$ generate $G\times C_p$. Indeed they generate a group projecting on both factors and hence if by contradiction it's a proper subgroup, this subgroup defines the graph of an isomorphism between some quotient of $G$ with $C_p$, contradiction.

2. for all $n\neq 2$, $\mathrm{SL}_n(\mathbf{Z})$ is perfect and in particular $\mathrm{Hom}(\mathrm{SL}_n(\mathbf{Z}),C_2)=0$. Proof: this is obvious for $n\le 1$ and for $n\ge 3$ it's generated by the elementary matrices $e_{ij}$, $i\neq j$, each of which being a commutator, namely $e_{ij}=[e_{ik},e_{kj}]$ for some $k\notin\{i,j\}$.

3. If $n$ is odd then $\mathrm{GL}_n(\mathbf{Z})\simeq\mathrm{SL}_n(\mathbf{Z})\times C_2$, where $C_2$ corresponds to $\{\pm I_n\}$.

Assuming the assertions your claim are true, it follows that $\mathrm{GL}_n(\mathbf{Z})$ is generated by 2 elements for all $n$; for $n$ odd (your missing case), the matrices $-s_1,s_3$ indeed generate $\mathrm{GL}_n(\mathbf{Z})$.

This follows from the following three facts:

1. Let $p$ be prime and $C_p=\langle c\rangle$ the cyclic group of order $p$. If $G$ is an arbitrary group with $\mathrm{Hom}(G,C_p)=0$, and if $G$ is generated by $k\ge 1$ elements, then so is $G\times C_p$. Proof: if $g_1,\dots,g_k$ generate $G$, then $(g_1,c),(g_2,1)\dots,(g_k,1)$ generate $G\times C_p$. Indeed they generate a group projecting onto both factors and hence if by contradiction it's a proper subgroup, this subgroup defines the graph of an isomorphism between some quotient of $G$ with $C_p$, contradiction.

2. for all $n\neq 2$, $\mathrm{SL}_n(\mathbf{Z})$ is perfect and in particular $\mathrm{Hom}(\mathrm{SL}_n(\mathbf{Z}),C_2)=0$. Proof: this is obvious for $n\le 1$ and for $n\ge 3$ it's generated by the elementary matrices $e_{ij}$, $i\neq j$, each of which being a commutator, namely $e_{ij}=[e_{ik},e_{kj}]$ for some $k\notin\{i,j\}$.

3. If $n$ is odd then $\mathrm{GL}_n(\mathbf{Z})\simeq\mathrm{SL}_n(\mathbf{Z})\times C_2$, where $C_2$ corresponds to $\{\pm I_n\}$.