Minimal polynomial, determinants and invertibility

Here is another way. Note that $m_A(x) \mid \mathrm{char}_A(x)$ in particular if $m_A(0)=0$ then $0$ is an eigenvalue of $A$. So $A$ has non-trivial kernel and is thereby not invertible.

Here's another method, if it helps:

We know that $m_A(A)=0,$ thus if $m_A(0)=0,$ i.e. if there is no constant term, then we can write $A^r-\lambda_{r-1}A^{r-1}+\cdots\pm\lambda_1 A=(A^{r-1}-\lambda_{r-1}A^{r-2}+\cdots\pm\lambda_1I)A=0.$ Since $A$ is invertible, multiplying on the right by $A^{-1}$ shows that $A$ satisfies a polynomial of lesser degree than its minimal polynomial, giving a contradiction.

You have a good shot, but there are two errors in your attempted proof.

1. In general, the constant term of the characteristic polynomial of $A$ is a (matrix independent) multiple of $\det A$, but the constant term of the minimal polynomial is not.
2. Even if we are talking about the characteristic polynomial, your sign of the constant term is not correct. The characteristic polynomial is $p_A(x)=\det(xI-A)$. Hence $p_A(0)=\det(-A)=(-1)^r\det A$. Although we don't really need the sign to answer this question here, it would be better if we get every detail right.

For a correct proof following your line of thought, see Jacob Schlather's answer. To learn a different perspective, you should read Andrew's answer too.