As referred to Why non-trivial solution only if determinant is zero,
that says "$(A−\lambda I)x=0$ has a nontrivial solution (a solution where $x\neq 0$) if and only if $\det(A−\lambda I)=0$ "
which I assume that can imply $(Ax=0)$ $\land$ $(x\neq 0)$ $\iff \det(A)=0$ for any square matrix $A$ and vector $x$
that can further imply $\det(A)=0 \implies (Ax=0) \land (x\neq 0)$ for any square matrix $A$ and vector $x$
But when $\begin{gather} A = \begin{bmatrix} 1 & 1 \\1 & 1 \end{bmatrix} \end{gather}$ which $\det(A)=0$, it doesn't imply $(Ax=0) \land (x\neq 0)$ for any vector $x$ because there exists a vector $x = \begin{bmatrix} 2 \\3 \end{bmatrix} ≠ \begin{bmatrix} 0 \\0 \end{bmatrix}$ but $Ax = \begin{bmatrix} 1 & 1 \\1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\3 \end{bmatrix} = \begin{bmatrix} 5 \\5 \end{bmatrix} ≠ \begin{bmatrix} 0 \\0 \end{bmatrix} $
So I am confused and wonder if it is actually "$(Ax=0)$ $\land$ $(x\neq 0)$ $\implies det(A)=0$" ?
It is a matter of quantifiers:
The statement is not $$\forall x \neq 0: Ax = 0 \Leftrightarrow \det A = 0$$
but rather $$\exists x \neq 0: Ax = 0 \Leftrightarrow \det A = 0 ~~.$$
In particular, this means that if $\det A = 0$, then there exists a vector $x \neq 0$ s.t. $Ax = 0$. Not just any vector; some vector.
In your case of $A := \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, the vector $x := \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ is not $0$, but $Ax = 0$.