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.