What is the smallest variety containing all monoids and all semigroups with a one-sided zero? What is the smallest variety (in the universal algebra sense) containing all monoids, all semigroups with a left zero, all semigroups with a right zero, and as few other models as possible? So far, the only thing I can come up with is the class of all semigroups with a distinguished idempotent element - is there some way to rule out more models using just equational laws?
 A: This extremely far from complete, but comments are not intended for (partial) answers.
So, we are working in the signature $\{\cdot,c\}$. I'll just give you an infinite family of identities that are satisfied by all three kinds of structures you ask for (other than $\cdot$ being associative).
For any finite word $\alpha$ in the alphabet $X\cup\{c\}$ (where $X$ is your set of variables), let $\bar\alpha$ the result of erasing occurrences of $c$ in $\alpha$. Then
$$
c\alpha c \approx c\bar\alpha c
$$
is such an identity. For example,
$$
c x c y c \approx c x y c,
$$
that is,
$$
\forall x,y\,( c\cdot x\cdot c\cdot y\cdot c =c\cdot x\cdot y\cdot c).
$$
For $\alpha=c\cdots c$ you recover a weak form of idempotency.
A: I believe I can expand Pedro Sánchez Terraf's partial answer to a full answer. Fix the signature $\{\cdot,c\}$, let $W$ be the variety of algebras generated by the monoids and semigroups with left or right $0$, and let $V$ be the variety of algebras axiomatized by the equations


*

*$(xy)z = x(yz)$

*$cc = c$

*$cxcyc = cxyc$
Let's call 3. the "Pedro equation". In fact, the full family of equations described in Pedro's answer follows already from these three equations, by using 2. to remove any interior occurrence of $c$ which is adjacent to one of the outer occurrences, and then using 3. to remove the remaining interior occurrences of $c$ one by one. So $V$ is finitely axiomatized.
Claim: $V = W$. 
Proof: Every monoid and every semigroup with left or right zero satisfies the equations defining $V$, so $W\subseteq V$. To show the converse, it suffices to show that every free algebra on finitely many generators in $V$ is in $W$. So let $F$ be free on the generators $\{x_1,\dots,x_n\}$. Let $M$, $L$, and $R$ be the free monoid, semigroup with left zero, and semigroup with right zero, respectively, on the generators $\{x_1,\dots,x_n\}$. Now since $M$, $L$, and $R$ are in $V$, there are natural maps $F\to M$, $F\to L$, and $F\to R$, defined in each case by $x_i\mapsto x_i$, and hence there is a map $\varphi\colon F\to M\times L\times R$. I claim that $\varphi$ is injective, so $F$ is isomorphic to a subalgebra of a product of algebras in $W$, and hence $F$ is in $W$.
It is not hard to give a normal form for elements of $F$: Given any term in $\{x_1,\dots,x_n\}$, associativity allows us to erase parentheses, idempotence of $c$ allows us to collapse adjacent occurrences of $c$, and Pedro's equation allows us to remove all occurrences of $c$ other than the "outermost two" (if there are at least three occurrences of $c$). So every element of $F$ is represented by a unique word of the form $w$ or $w_l\cdot c\cdot w_r$ or $w_l\cdot c\cdot w\cdot c\cdot w_r$, where $w$ is a nonempty word in the generators, and $w_l,w_r$ are possibly empty words in the generators. Now a case analysis shows that if $t_1\neq t_2$ are distinct words, each of one of the three forms above, then $t_1$ and $t_2$ can be separated by one of the maps to $M$, $L$, or $R$, and hence they are separated by $\varphi$.
For example, if $t_1$ has the form $w_l\cdot c\cdot w\cdot c\cdot w_r$ and $t_2$ has the form $w_l'\cdot c\cdot w_r'$, then $\varphi(t_1) = (w_l\cdot w\cdot w_r,w_l\cdot c,c\cdot w_r)$ and $\varphi(t_2) = (w_l'\cdot w_r',w_l'\cdot c,c\cdot w_r')$. Then,


*

*If $w_l \neq w_l'$, then $w_l\cdot c \neq w_l'\cdot c$ in $L$.

*If $w_r \neq w_r'$, then $c\cdot w_r\neq c\cdot w_r'$ in $R$.

*If $w_l = w_l'$ and $w_r = w_r'$, then since $w$ is nonempty, $w_l\cdot w\cdot w_r\neq w_l'\cdot w_r'$ in $M$.

