This question was edited on March 12, 2023 and I was asked to comment on the edited form. Let me copy the essential part of the new question:

I believe $\ldots$ that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

The if part of this statement is correct, while the only if part is not correct.

(if holds):
Assume that the set $\Sigma$ of subdirectly irreducible quotients of the initial algebra $\mathbf{F}_0$ of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism. This means that $\Sigma$ contains an isomorphic copy of every subdirectly irreducible algebra in $\mathscr{V}$. We have $\mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. The first equality holds because every algebra in $\mathscr{V}$ is a subdirect product of subdirectly irreducible members of $\mathscr{V}$ and $\Sigma$ contains a subdirectly irreducible of every isomorphism type. For the second equality, $\subseteq$ holds because $\Sigma\subseteq\mathsf{H}(\mathbf{F}_0)$, while $\supseteq$ holds because $\mathbf{F}_0\in \mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)$.

(only if fails):
The simplest counterexample is the variety $\mathscr{V}$ of all algebras in a language with two constant symbols $a, b$, and no other operations. The algebras $\mathbf A\in \mathscr{V}$ are just structures $\mathbf A = \langle A; a, b\rangle$ where $a^{\mathbf A}, b^{\mathbf A}\in A$. We allow $a^{\mathbf A}=b^{\mathbf A}$. In this example, $\mathbf{F}_0$ is (up to isomorphism) the algebra $\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{F}_0}=0$ and $b^{\mathbf{F}_0}=1$. It is the case that $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. In this example, $\mathscr{V}$ has two isomorphism types of subdirectly irreducible algebras. $\mathbf{F}_0$ is itself subdirectly irreducible, but there is another one: $\mathbf{E}=\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{E}}=0=b^{\mathbf{E}}$. This second one $\mathbf{E}$ is not generated by the empty set. To connect this back to the statement of the question, $\mathscr{V}$ is generated by its initial algebra but it is not the case that all subdirectly irreducible members of $\mathscr{V}$ are quotients of the initial algebra. ($\mathbf{E}$ is not.)

A more serious reason why only if fails:
If $\mathscr{V}$ has the property that all subdirectly irreducibles are quotients of the initial object, then the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set. (There is a set containing at least one isomorphic copy of each subdirectly irreducible algebra in $\mathscr{V}$.) We say that $\mathscr{V}$ is residually small when the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set and we say that $\mathscr{V}$ is residually large when the class of subdirectly irreducibles of $\mathscr{V}$ cannot be represented by a set. So, the property that all subdirectly irreducibles are quotients of the initial object forces $\mathscr{V}$ to be residually small. But the property $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$ does not force $\mathscr{V}$ to be residually small. Take any finite, nilpotent, nonabelian group $G$, and let $G_G$ denote its expansion by constants. Let $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(G_G)$. In this example $G_G$ is the initial object of $\mathscr{V}$ (and $\mathscr{V}$ is generated by this initial object), but $\mathscr{V}$ is residually large. Thus, $\mathscr{V}$ is generated by its initial object, but the class of subdirectly irreducible members does not coincide with the class of subdirectly irreducible quotients of the initial object.

This question was edited on March 12, 2023 and I was asked to comment on the edited form. Let me copy the essential part of the new question:

I believe $\ldots$ that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

The if part of this statement is correct, while the only if part is not correct.

(if holds):
Assume that the set $\Sigma$ of subdirectly irreducible quotients of the initial algebra $\mathbf{F}_0$ of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism. This means that $\Sigma$ contains an isomorphic copy of every subdirectly irreducible algebra in $\mathscr{V}$. We have $\mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. The first equality holds because every algebra in $\mathscr{V}$ is a subdirect product of subdirectly irreducible members of $\mathscr{V}$ and $\Sigma$ contains a subdirectly irreducible of every isomorphism type. For the second equality, $\subseteq$ holds because $\Sigma\subseteq\mathsf{H}(\mathbf{F}_0)$, while $\supseteq$ holds because $\mathbf{F}_0\in \mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)$.

(only if fails):
The simplest counterexample is the variety $\mathscr{V}$ of all algebras in a language with two constant symbols $a, b$, and no other operations. The algebras $\mathbf A\in \mathscr{V}$ are just structures $\mathbf A = \langle A; a, b\rangle$ where $a^{\mathbf A}, b^{\mathbf A}\in A$. We allow $a^{\mathbf A}=b^{\mathbf A}$. In this example, $\mathbf{F}_0$ is (up to isomorphism) the algebra $\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{F}_0}=0$ and $b^{\mathbf{F}_0}=1$. It is the case that $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. In this example, $\mathscr{V}$ has two isomorphism types of subdirectly irreducible algebras. $\mathbf{F}_0$ is itself subdirectly irreducible, but there is another one: $\mathbf{E}=\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{E}}=0=b^{\mathbf{E}}$. This second one, $\mathbf{E}$, is not generated by the empty set. To connect this back to the statement of the question, $\mathscr{V}$ is generated by its initial algebra but it is not the case that all subdirectly irreducible members of $\mathscr{V}$ are quotients of the initial algebra. ($\mathbf{E}$ is not.)

A more serious reason why only if fails:
If $\mathscr{V}$ has the property that all subdirectly irreducibles are quotients of the initial object, then the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set. (There is a set containing at least one isomorphic copy of each subdirectly irreducible algebra in $\mathscr{V}$.) We say that $\mathscr{V}$ is residually small when the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set and we say that $\mathscr{V}$ is residually large when the class of subdirectly irreducibles of $\mathscr{V}$ cannot be represented by a set. So, the property that all subdirectly irreducibles are quotients of the initial object forces $\mathscr{V}$ to be residually small. But the property $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$ does not force $\mathscr{V}$ to be residually small. Take any finite, nilpotent, nonabelian group $G$, and let $G_G$ denote its expansion by constants. Let $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(G_G)$. In this example $G_G$ is the initial object of $\mathscr{V}$ (and $\mathscr{V}$ is generated by this initial object), but $\mathscr{V}$ is residually large. Thus, $\mathscr{V}$ is generated by its initial object, but the class of subdirectly irreducible members does not coincide with the class of subdirectly irreducible quotients of the initial object.

This question was edited on March 12, 2023 and I was asked to comment on the edited form. Let me copy the essential part of the new question:

I believe $\ldots$ that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

The if part of this statement is correct, while the only if part is not correct.

(if holds):
Assume that the set $\Sigma$ of subdirectly irreducible quotients of the initial algebra $\mathbf{F}_0$ of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism. This means that $\Sigma$ contains an isomorphic copy of every subdirectly irreducible algebra in $\mathscr{V}$. We have $\mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. The first equality holds because every algebra in $\mathscr{V}$ is a subdirect product of subdirectly irreducible members of $\mathscr{V}$ and $\Sigma$ contains a subdirectly irreducible of every isomorphism type. For the second equality, $\subseteq$ holds because $\Sigma\subseteq\mathsf{H}(\mathbf{F}_0)$, while $\supseteq$ holds because $\mathbf{F}_0\in \mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)$.

(only if fails):
The simplest counterexample is the variety $\mathscr{V}$ of all algebras in a language with two constant symbols $a, b$, and no other operations. The algebras $\mathbf A\in \mathscr{V}$ are just structures $\mathbf A = \langle A; a, b\rangle$ where $a^{\mathbf A}, b^{\mathbf A}\in A$. We allow $a^{\mathbf A}=b^{\mathbf A}$. In this example, $\mathbf{F}_0$ is (up to isomorphism) the algebra $\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{F}_0}=0$ and $b^{\mathbf{F}_0}=1$. It is the case that $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. In this example, $\mathscr{V}$ has two isomorphism types of subdirectly irreducible algebras. $\mathbf{F}_0$ is itself subdirectly irreducible, but there is another one: $\mathbf{E}=\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{E}}=0=b^{\mathbf{E}}$. This second one $\mathbf{E}$ is not generated by the empty set. To connect this back to the statement of the question, $\mathscr{V}$ is generated by its initial algebra but it is not the case that all subdirectly irreducible members of $\mathscr{V}$ are quotients of the initial algebra. ($\mathbf{E}$ is not.)

A more serious reason why only if fails:
If $\mathscr{V}$ has the property that all subdirectly irreducibles are quotients of the initial object, then the class of subdirectly irreducibles of $\mathscr{V}$ must be a set rather than a proper class. We say that $\mathscr{V}$ is residually small when the class of subdirectly irreducibles of $\mathscr{V}$ is a set and we say that $\mathscr{V}$ is residually large when the class of subdirectly irreducibles of $\mathscr{V}$ is a proper class. So, the property that all subdirectly irreducibles are quotients of the initial object forces $\mathscr{V}$ to be residually small. But the property $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$ does not force $\mathscr{V}$ to be residually small. Take any finite, nilpotent, nonabelian group $G$, and let $G_G$ denote its expansion by constants. Let $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(G_G)$. In this example $G_G$ is the initial object of $\mathscr{V}$ (and $\mathscr{V}$ is generated by this initial object), but $\mathscr{V}$ is residually large. Thus, $\mathscr{V}$ is generated by its initial object, but the class of subdirectly irreducible members does not coincide with the class of subdirectly irreducible quotients of the initial object.

This question was edited on March 12, 2023 and I was asked to comment on the edited form. Let me copy the essential part of the new question:

I believe $\ldots$ that a variety $\mathscr V$ is of the above kind if and only if all of its subdirectly irreducible algebras can be generated by no elements. That is, subdirectly irreducible quotients of the initial algebra of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism.

The if part of this statement is correct, while the only if part is not correct.

(if holds):
Assume that the set $\Sigma$ of subdirectly irreducible quotients of the initial algebra $\mathbf{F}_0$ of $\mathscr V$ are the only subdirectly irreducibles in $\mathscr V$ up to isomorphism. This means that $\Sigma$ contains an isomorphic copy of every subdirectly irreducible algebra in $\mathscr{V}$. We have $\mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. The first equality holds because every algebra in $\mathscr{V}$ is a subdirect product of subdirectly irreducible members of $\mathscr{V}$ and $\Sigma$ contains a subdirectly irreducible of every isomorphism type. For the second equality, $\subseteq$ holds because $\Sigma\subseteq\mathsf{H}(\mathbf{F}_0)$, while $\supseteq$ holds because $\mathbf{F}_0\in \mathscr{V}=\mathsf{S}\mathsf{P}(\Sigma)$.

(only if fails):
The simplest counterexample is the variety $\mathscr{V}$ of all algebras in a language with two constant symbols $a, b$, and no other operations. The algebras $\mathbf A\in \mathscr{V}$ are just structures $\mathbf A = \langle A; a, b\rangle$ where $a^{\mathbf A}, b^{\mathbf A}\in A$. We allow $a^{\mathbf A}=b^{\mathbf A}$. In this example, $\mathbf{F}_0$ is (up to isomorphism) the algebra $\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{F}_0}=0$ and $b^{\mathbf{F}_0}=1$. It is the case that $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$. In this example, $\mathscr{V}$ has two isomorphism types of subdirectly irreducible algebras. $\mathbf{F}_0$ is itself subdirectly irreducible, but there is another one: $\mathbf{E}=\langle \{0,1\}; a, b\rangle$ with $a^{\mathbf{E}}=0=b^{\mathbf{E}}$. This second one $\mathbf{E}$ is not generated by the empty set. To connect this back to the statement of the question, $\mathscr{V}$ is generated by its initial algebra but it is not the case that all subdirectly irreducible members of $\mathscr{V}$ are quotients of the initial algebra. ($\mathbf{E}$ is not.)

A more serious reason why only if fails:
If $\mathscr{V}$ has the property that all subdirectly irreducibles are quotients of the initial object, then the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set. (There is a set containing at least one isomorphic copy of each subdirectly irreducible algebra in $\mathscr{V}$.) We say that $\mathscr{V}$ is residually small when the class of subdirectly irreducibles of $\mathscr{V}$ is represented by a set and we say that $\mathscr{V}$ is residually large when the class of subdirectly irreducibles of $\mathscr{V}$ cannot be represented by a set. So, the property that all subdirectly irreducibles are quotients of the initial object forces $\mathscr{V}$ to be residually small. But the property $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(\mathbf{F}_0)$ does not force $\mathscr{V}$ to be residually small. Take any finite, nilpotent, nonabelian group $G$, and let $G_G$ denote its expansion by constants. Let $\mathscr{V}=\mathsf{H}\mathsf{S}\mathsf{P}(G_G)$. In this example $G_G$ is the initial object of $\mathscr{V}$ (and $\mathscr{V}$ is generated by this initial object), but $\mathscr{V}$ is residually large. Thus, $\mathscr{V}$ is generated by its initial object, but the class of subdirectly irreducible members does not coincide with the class of subdirectly irreducible quotients of the initial object.