Your preliminary statement needs clarification: a linear continuous operator $T$ from $\mathscr S(\mathbb R^n)$ into $\mathscr S'(\mathbb R^n)$ has a kernel $k$ in $\mathscr S'(\mathbb R^n\times\mathbb R^n )$ so that for $\phi, \psi\in \mathscr S(\mathbb R^n)$,
$$
\langle T\psi,\phi\rangle_{\mathscr S'(\mathbb R^n),\mathscr S(\mathbb R^n)}
=\langle k(x,y),\phi(x) \psi (y)\rangle_{\mathscr S'(\mathbb R^{2n}),\mathscr S'(\mathbb R^{2n})}.
$$
So some operators may have a very singular kernel everywhere and not only on the diagonal. A simple example of an operator with a singular kernel off diagonal is the translation of vector $h$:
$$
(T\phi)(x)=\phi(x-h),
$$
whose kernel is $k(x,y)=\delta_0(x-h-y)$, singular at $x=h+y$.

Your statement (1) is incorrect: the identity has the kernel $\delta_0(x-y)$ and this answers negatively (2).

What is true about pseudodifferential operators is that their kernels are indeed smooth off diagonal but more and more singular at the diagonal when you increase the order of the operator. The simplest example is to consider, say in one dimension, the operator $(d/dx)^N$, a differential operator of order $N$: its kernel is
$$
\delta_0^{(N)}(x-y).
$$
Operators with kernels in $\mathscr S(\mathbb R^{2n})$ are some particular type of regularizing operators.