It's true that if a matrix is singular then any equation with at least one solution has infinitely many. Here is one way to see this.
Let $A$ be a singular matrix, so that $Ax=0$ has a nonzero solution $x_0$. Since $A$ is linear, any multiple of $x_0$ is also a solution, as mentioned in a comment. Thus we know at least there are infinitely many solutions to $Ax=0$. The set of such solutions is called the kernel of the matrix, and it is actually a vector subspace. This is easy to see; if $Ax=0$ and $Ay=0$ and $a,b$ are scalars, then
$$A(ax+by)=aAx+bAy=0$$
so any linear combination of two elements of the kernel is still in the kernel.
Let $K$ be the kernel of $A$. Let's say we want to solve
$$Ax=y$$
If a solution exists, we can find at least one by methods you're probably familiar with. Call some solution $x_0$. How do we get the rest of the solutions? Well, if we add an element $k$ of $K$ to $x_0$ we get
$$A(x_0+k)=Ax_0+Ak=y+0=y$$
Hence $x_0+k$ is also a solution.
It turns out that this in fact gives you every solution. Why is that? Again this is because of linearity. Suppose $x_0$ and $x_1$ are both solutions. Then
$$A(x_1-x_0)=Ax_1-Ax_0=y-y=0$$
Hence $x_1-x_0\in K$, and $x_1=x_0+(x_1-x_0)$.
You want to prove that (a linear system of equtions is inconsistent) implies that (there is a pivot column in the last column of an echelon form of the augmented matrix).
This is equivalent to showing that
(there is no pivot column in the last column of the echelon form of the augmented matrix) implies that (the linear system is consistent).
Now suppose that there is no pivot column of the echelon form of the augmented system. Suppose that you already know that the solution space is not changed by elementary row operations. Also note that dropping the zero rows do not change the solution space as well and hence we can assume that there is no zero rows in our RREF. Remark: if the RREF is the zero matrix, any vector of the right size is a solution, we will assume that this is not the case here.
Let's construct a solution $x$ that satisfies the linear system. WLOG, we can write $x=(x_B, x_N)$ where $x_B$ are the entries of $x$ corresponds to pivot columns and $x_N$ corresponds to entries of $x$ corresponds to non-pivot columns.
We want to solve
$$Rx=r$$
$$\begin{bmatrix} R_B & R_N \end{bmatrix}\begin{bmatrix} x_B \\ x_N \end{bmatrix}=r$$
where $R_B$ corresponds to the pivot columns and $R_N$ corresponds to the non-zero columns.
We set $x_N=0$ and we obtain $$R_Bx_B=r$$
As, $R_B$ is constructed from the pivot columns, the diagonal entries of $R_B$ are non-zero, hence it is non-singular and hence $x_B= R_B^{-1}r$.
Hence, we have found a solution.
Best Answer
As I interpret it, they mean by "augmented matrix" the full coefficient matrix, where the equation $$ \cases{ax + by = c\\dx+ey=f} $$ gets rewritten to the matrix $$ \begin{bmatrix}a&b&c\\d&e&f\end{bmatrix} $$ I hope you can see that if this matrix has a pivot in the right-most column, then the system has no solution.