The answer is yes since a fiberwise condition (such as affine stabilizers) does not imply a global condition (such as affine diagonal) without extra hypotheses (such as having the resolution property). Think of quasi-finite+proper <=> finite.

There are (non-separated) schemes with non-affine diagonal, for example, two copies of the affine plane glued together outside the origin.

Let me instead pose the following

Challenge: Find a stack X with affine stabilizers with a diagonal which is separated but not quasi-affine.

Remark 1: Note that if X has finite stabilizers then the diagonal is quasi-finite and separated, hence quasi-affine (Zariski's MT).

Remark 2: It we drop the condition that the diagonal of X is separated, it is easy to find examples.

Remark 3: The stabilizers of X are affine if and only if they are quasi-affine.

(throughout "stabilizers of X" means stabilizers of points of X)

The answer is yes since a fiberwise condition (such as affine stabilizers) does not imply a global condition (such as affine diagonal) without extra hypotheses (such as having the resolution property). Think of quasi-finite+proper <=> finite.

There are (non-separated) schemes with non-affine diagonal, for example, two copies of the affine plane glued together outside the origin.

Also see the related question.

The answer is no since a fiberwise condition (such as affine stabilizers) does not imply a global condition (such as affine diagonal) without extra hypotheses (such as having the resolution property). Think of quasi-finite+proper <=> finite.

For a counter-example, you can take a (non-separated) scheme with non-affine diagonal. For example, two copies of the affine plane glued together outside the origin.

Let me instead pose the following

Challenge: Find a stack X with affine stabilizers with a diagonal which is separated but not quasi-affine.

Remark 1: Note that if X has finite stabilizers then the diagonal is quasi-finite and separated, hence quasi-affine (Zariski's MT).

Remark 2: It we drop the condition that the diagonal of X is separated, it is easy to find examples.

Remark 3: The stabilizers of X are affine if and only if they are quasi-affine.

(throughout "stabilizers of X" means stabilizers of points of X)

The answer is yes since a fiberwise condition (such as affine stabilizers) does not imply a global condition (such as affine diagonal) without extra hypotheses (such as having the resolution property). Think of quasi-finite+proper <=> finite.

There are (non-separated) schemes with non-affine diagonal, for example, two copies of the affine plane glued together outside the origin.

Let me instead pose the following

Challenge: Find a stack X with affine stabilizers with a diagonal which is separated but not quasi-affine.

Remark 1: Note that if X has finite stabilizers then the diagonal is quasi-finite and separated, hence quasi-affine (Zariski's MT).

Remark 2: It we drop the condition that the diagonal of X is separated, it is easy to find examples.

Remark 3: The stabilizers of X are affine if and only if they are quasi-affine.

(throughout "stabilizers of X" means stabilizers of points of X)