I think the easiest answer is the bicyclic monoid $B$ given by the presentation $\langle a,b\mid ab=1\rangle$. Considering a unilateral shift on Hilbert space and its adjoint show that $ba\ne 1$. But of course $ab=1\implies ba=1$ in a finite monoid so $ba$ and $1$ are inseparable in a finite monoid. Actually, it is known that any proper image of $B$ is a cyclic group. There are no invertible elements in $B$ since using the relation, it is easy to see that every element is of the form $b^ma^k$ with $m,k$ nonnegative integers. If the element is invertible and not the identity, then either $a$ would have a left inverse or $b$ would have a right inverse, which would make $B$ a group.

Actually, it is known that $B$ cannot be embedded in any compact Hausdorff topological monoid. In particular, it cannot embed in its profinite completion.

If one takes the monoid with presentation $\langle a,b,c,d\mid ab=1=cd, cb=0=ad\rangle$, where 0 is a multiplicative zero, then one gets a monoid with no nontrivial proper images and trivial group of units, but this is a little harder to prove.

I think the easiest answer is the bicyclic monoid $B$ given by the presentation $\langle a,b\mid ab=1\rangle$. Considering a unilateral shift on Hilbert space and its adjoint show that $ba\ne 1$. But of course $ab=1\implies ba=1$ in a finite monoid so $ba$ and $1$ are inseparable in a finite monoid. Actually, it is known that any proper image of $B$ is a cyclic group. There are no invertible elements in $B$ since using the relation, it is easy to see that every element is of the form $b^ma^k$ with $m,k$ nonnegative integers. If the element is invertible and not the identity, then either $a$ would have a left inverse or $b$ would have a right inverse, which would make $B$ a group.

Actually, it is known that $B$ cannot be embedded in any compact Hausdorff topological monoid. In particular, it cannot embed in its profinite completion.

If one takes the monoid (with zero) with presentation $\langle a,b,c,d\mid ab=1=cd, cb=0=ad\rangle$ in that category, where 0 is a multiplicative zero, then one gets a monoid with no nontrivial proper images and trivial group of units, but this is a little harder to prove.

I think the easiest answer is the bicyclic monoid $B$ given by the presentation $\langle a,b\mid ab=1\rangle$. Considering a unilateral shift on Hilbert space and its adjoint show that $ba\ne 1$. But of course $ab=1\implies ba=1$ in a finite monoid so $ba$ and $1$ are inseparable in a finite monoid. Actually, it is known that any proper image of $B$ is a cyclic group. There are no invertible elements in $B$ since using the relation, it is easy to see that every element is of the form $b^ma^k$ with $m,k$ nonnegative integers. If the element is invertible and not the identity, then either $a$ would have a left inverse or $b$ would have a right inverse, which would make $B$ a group.

Actually, it is known that $B$ cannot be embedded in any compact Hausdorff topological monoid. In particular, it cannot embed in it's profinite completion.

If one takes the monoid with presentation $\langle a,b,c,d\mid ab=1=cd, cb=0=ad\rangle$, where 0 is a multiplicative zero, then one gets a monoid with no nontrivial proper images and trivial group of units, but this is a little harder to prove.

I think the easiest answer is the bicyclic monoid $B$ given by the presentation $\langle a,b\mid ab=1\rangle$. Considering a unilateral shift on Hilbert space and its adjoint show that $ba\ne 1$. But of course $ab=1\implies ba=1$ in a finite monoid so $ba$ and $1$ are inseparable in a finite monoid. Actually, it is known that any proper image of $B$ is a cyclic group. There are no invertible elements in $B$ since using the relation, it is easy to see that every element is of the form $b^ma^k$ with $m,k$ nonnegative integers. If the element is invertible and not the identity, then either $a$ would have a left inverse or $b$ would have a right inverse, which would make $B$ a group.

Actually, it is known that $B$ cannot be embedded in any compact Hausdorff topological monoid. In particular, it cannot embed in its profinite completion.

If one takes the monoid with presentation $\langle a,b,c,d\mid ab=1=cd, cb=0=ad\rangle$, where 0 is a multiplicative zero, then one gets a monoid with no nontrivial proper images and trivial group of units, but this is a little harder to prove.