I will explain why
any collection of isomorphism types of
subgroups of symmetric groups can be realized
as $\mathbb G(A)$ for some algebra $A$.
<p>
But first, let me address this phrase from the problem statement:
<b><i>in which each variable actually appears</i></b>.
This condition does not restrict.
If an algebra $A$ has a term $t(x,y,z)$ where only $x$ and $y$
`actually appear', then one can expand $A$ by adding
a new fundamental operation $f$ interpreted so that
$f(x,y,z) := t(x,y,z)$ on $A$. This has no effect on
which functions are the interpretations
of $n$-ary terms for $A$,
but it does have an effect on whether variables `actually
appear' in the term: each of $x$, $y$, and $z$
actually appears in $f$, while $f$ has the same
symmetry group as $t$. 
<br>

Instead of the condition on term operations
<b><i>in which each variable actually appears</i></b>
I shall work with the condition on term operations
<b><i>which depend on all variables</i></b>.
<p>

Now, back to the problem.
I claimed above that any collection of isomorphism types of
subgroups of symmetric groups can be realized
as $\mathbb G(A)$ for some algebra $A$. This is almost true.
You must include the isomorphism type of the $1$-element
group, since it is the symmetry group of $t(x)=x$.
So let $\mathcal K$ be any class of subgroups of finite
symmetric groups which includes the $1$-element group. 
For a given $G\in\mathcal K$,
where $G\leq S_n$,
define an $n$-ary operation $f_G(x_1,\ldots,x_n)$ on the set
$\{a, b\}\cup \mathbb N^+
= \{a, b, 1, 2, 3, \ldots\}$ as follows:
$$
f_G(\bar{u}) = \begin{cases} a & \textrm{if $\bar{u} = 
(\sigma(1),\ldots,\sigma(n))$ for some $\sigma\in G$;}\\
b & \textrm{otherwise.}
 \end{cases}
$$
Let $A$ be the algebra on $\{a,b\}\cup \mathbb N^+$
whose operations are all operations
of the form $f_G$, $G\in\mathcal K$.
<p>

I claim that the following are true.<br>
* Each $f_G$ depends on all of its variables.<br>
* The symmetry group of $f_G(x_1,\ldots,x_n)$ is $G\;(\leq S_n)$.<br>
* Any term operation of $A$ that depends on all of its
variables is obtained from one of the $f_G$'s by
permuting the variables. Hence, its symmetry
group is conjugate to $G$ in $S_n$.<br>

This is enough to show that $\mathbb G(A)$
consists of exactly the isomorphism
types
of groups in $\mathcal K$.<p>

**Edit.** Noah didn't like me changing his question, so let me say what the above construction contributes to the problem in its original form.<p>

If $f_G(x_1,\ldots,x_n)$ is defined as above, and we identify two variables: $f_G(\underline{x_2},\underline{x_2},x_3,x_4,\ldots,x_n)$, then this operation is constant and its variable-set is $\{x_2,\ldots,x_n\}$.
Its symmetry group in Noah's sense is $S_{n-1}$. Similarly, the composite $f_G(f_G(x_1,\ldots,x_n),y_2,\ldots,y_n)$ is constant and its variable-set is $\{x_1,\ldots,x_n,y_2,\ldots,y_n\}$, so its symmetry group is $S_{2n-1}$. Using tricks like this one can obtain every finite symmetric group as a symmetry group of a term of $A$, as long as $A$ has at least one operation $f_G$ of arity greater than $1$. 
The isomorphism types of Noah-symmetry groups turn out to be 
exactly those in $\mathcal K\cup \{S_n\;|\;n=1, 2, \ldots\}$.