(Remark: Sam Roberts pointed out to me an error in the first version of the proof I posted earlier -- there was no reason that the desired automorphisms exist, as there was no specification of them on the atoms outside of the sets of atoms under consideration. The following is a significantly modified version, and I think this time it works. One should also compare with Emil Jeřábek's and Sam Roberts' comments regarding DC$_{\omega_1}$ in the original question and below this answer.)
Assuming foundation is included (in the sense that for every non-empty set $x$ with no atomic elements, there is $y\in x$ such that $y\cap x=\emptyset$), SDC follows:
Write $U$ for the entire universe (of all sets and atoms).
If there is only a set $A$ of atoms, then $U=V(A)$, i.e. $U$ is the union of the cumulative hierarchy $\left<V_\beta(A)\right>_{\beta\in\mathrm{Ord}}$ above $V_0(A)=A$. In this case there's an easy argument by recursively collecting witnesses according to rank. That is, let $\beta_0$ be the least ordinal $\beta$ such that there is $y\in V_\beta(A)$ with $\varphi(z,y)$ (where $z$ was the given parameter). Then by collection, we can take $\beta_1$ least such that for all $x\in V_{\beta_0}(A)$ there is $y\in V_{\beta_1}(A)$ such that $\varphi(x,y)$. Etc, producing $\left<\beta_n\right>_{n<\omega}$. From this information we can build a set relation and apply DC to it to get the DC-branch $f$, i.e. such that $f(0)=z$ and $\varphi(f(n),f(n+1))$ for each $n<\omega$.
So suppose $A$ is a proper class. We will adapt the argument just mentioned.
Remark: I use parametrized collection in the proof, but this follows from the axioms in a routine way, by folding parameters into the ``$x$'' in unparametrized collection and modifying $\phi$ appropriately.
Lemma 1: For every set $x$, the transitive closure of $x$ exists.
Proof: Just the usual thing, taking the union of $\bigcup^nx$ over all $n<\omega$.
Given a set of atoms $a$ and ordinal $\beta$, let $V_\beta(a)$ be the cumulative hierarchy built above $V_0(a)=a$.
Lemma 2: For every set $x$, there is a set $a$ of atoms and ordinal $\beta$ with $x\in V_\beta(a)$.
Proof: Let $t$ be the transitive closure of $\{x\}$ and $a=t\cap A$. So it suffices to see $t\subseteq V(a)$. If not let $t'=t\backslash V(a)$, so $t'\neq\emptyset$ and $t'$ is not an atom, and note that $t'$ contains no atoms. By foundation there is (a set) $y\in t'$ such that $y\cap t'=\emptyset$. So since $y\subseteq t$, we have $y\subseteq V(a)$. But then by collection there is $\beta$ with $y\subseteq V_\beta(a)$, but then $y\in V_{\beta+1}(a)$, so $y\in V(a)$, a contradiction.
We now split into cases.
Case 1: For all cardinals $\kappa$ there is a set of atoms of cardinality $\kappa$.
Claim 1.1: Let $a,a'$ be sets of atoms of the same cardinality and $\pi:a\to a'$ a bijection, and $\pi^+:V(a)\to V(a')$ the resulting isomorphism. Then for all formulas $\varphi$ and all $x\in V(a)$, we have $\varphi(x)\Leftrightarrow\varphi(\pi^+(x))$ (with truth of $\varphi$ evaluated in $U$).
Proof: By (meta-)induction on formula complexity. If $\varphi$ is $\Sigma_0$ then it is just absoluteness of $\Sigma_0$ between $V(a)$ and $U$ and since $\pi^+$ is an isomorphism (the language can include a predicate interpreted as $A$, which is equivalent to the sets $a/a'$ when interpreted over $V(a)/V(a')$). Suppose it holds for $\Sigma_n$ formulas, let $\varphi$ be $\Sigma_n$, let $x\in V(a)$, and suppose there is $y\in U$ such that $\varphi(x,y)$. Let $b$ be some set of atoms which is disjoint from $a$ and such that there is such a $y\in V(a\cup b)$. Let $b'$ be a set of atoms of the same cardinality as $b$, which is disjoint from $a'$. This exists by Case 1 hypothesis. Then $a'\cup b'$ has the same cardinality as $a\cup b$. Let $\sigma:a\cup b\to a'\cup b'$ be a bijection extending $\pi$ (which bijected $a$ with $a'$ already). Let $\sigma^+:V(a\cup b)\to V(a'\cup b')$ be the resulting isomorphism. Then $\pi^+\subseteq\sigma^+$, and by induction, since $\varphi(x,y)$ is true, so is $\varphi(\sigma^+(x),\sigma^+(y))$, which suffices, proving Claim 1.1.
Now suppose that for all $x$ there is $y$ such that $\varphi(x,y)$, and let $z$ be given. Let $t_z$ be the transitive closure of $\{z\}$ and $a_z=A\cap t_z$. Let $(\kappa_0,\beta_0)$ be the lexicographically least pair $(\kappa,\beta)$ of ordinals such that for some set $a$ of atoms of cardinality $\kappa$, with $a$ disjoint from $a_z$, there is $y\in V_\beta(a_z\cup a)$ such that $\varphi(z,y)$.
Now suppose we have defined $(\kappa_0,\beta_0),\ldots,(\kappa_n,\beta_n)$.
Let $(\kappa_{n+1},\beta_{n+1})$ be the lex least pair $(\kappa,\beta)$ of ordinals such that for all sets $a_0,a_1,\ldots,a_n$ of atoms, if $a_z,a_0,\ldots,a_n$ are pairwise disjoint and each $a_i$ has cardinality $\kappa_i$, then there is a set $a'$ of atoms of cardinality $\kappa$ such that $a'$ is disjoint from $a_z\cup a_0\cup\ldots\cup a_n$, and for all $x\in V_{\beta_n}(a_z\cup a_0\cup\ldots\cup a_n)$, there is $y\in V_{\beta}(a_z\cup a_0\cup\ldots\cup a_n\cup a')$ such that $\varphi(x,y)$.
(Note that we allow $\kappa_n=0$.)
(Such a pair $(\kappa,\beta)$ exists. For let $(a_0,\ldots,a_n)$ and $(a'_0,\ldots,a'_n)$ satisfy these conditions.
Using collection, there is a set $Y$ such that for all $x\in W=V_{\beta_n}(a_z\cup a_0\cup\ldots\cup a_n)$, there is $y\in Y$ such that $\varphi(x,y)$. From $Y$ we can compute a $(\kappa,\beta)$ which works for $W$. But then the same $(\kappa,\beta)$ works for $W'=V_{\beta_n}(a_z\cup a'_0\cup\ldots\cup a'_n)$, by Claim 1.1 and since by case hypothesis, we can always find a disjoint set of atoms of a given cardinality.)
Now fix a sequence $\left<a_n\right>_{n<\omega}$ of pairwise disjoint sets of atoms, disjoint from $a_z$, with each $a_i$ of cardinality $\kappa_i$. (Exists by case hypothesis.) Note that there is $y\in V_{\beta_0}(a_z\cup a_0)$ such that $\varphi(z,y)$ and for each $n<\omega$, for each $x\in V_{\beta_n}(a_z\cup a_0\cup\ldots\cup a_n)$ there is $y\in V_{\beta_{n+1}}(a_z\cup a_0\cup\ldots\cup a_{n+1})$ such that $\varphi(x,y)$. So we have reduced the whole issue down to set-size, so applying choice, we get our DC-branch $f$, as desired.
Case 2: Otherwise. (There is some cardinal $\kappa$ such that there is no set of atoms of cardinality $\kappa$.)
Note that there are infinite sets of atoms, because for each $n<\omega$ there is a set of atoms of cardinality $n$, and applying collection. And given any set $a$ of atoms, there is an infinite set $b$ of atoms which is disjoint from $a$, for similar reasons and because $A$ is not a set.
Claim 2.1: There is a largest cardinal $\kappa$ such that there is a set of atoms of size $\kappa$.
Proof: Let $\kappa$ be the sup of all $\kappa'$ such that there is a set of atoms of size $\kappa'$. So $\kappa\geq\omega$. By collection, we can find a set $Y$ which contains a set of atoms of size $\kappa'$, for each $\kappa'<\kappa$. Let $\kappa$ be the cardinality the set of atoms in the transitive closure of $Y$. Then $\kappa$ works.
Let $\kappa_0=$ this largest $\kappa$.
Subcase 2.1: For every set $c$ of atoms of cardinality $\kappa_0$, there is a set $b$ of atoms of cardinality $\kappa_0$ which is disjoint from $c$.
This subcase is just a simplification of the subcase below.
Subcase 2.2: Otherwise. (There is a set $c$ of atoms of cardinality $\kappa_0$ such that there is no set $b$ of atoms of cardinality $\kappa_0$ with $c\cap b=\emptyset$.)
Let $c$ be a set as in Subcase 2.2 hypothesis. Then define $\mu_c$ as the largest cardinality of a set of atoms $b$ which is disjoint from $c$. This largest cardinality exists like in the proof of Claim 2.1. Now let $\mu_0$ be the least value of $\mu_c$, ranging over all such $c$.
Let $c_0$ witness this choice. Note that $\mu_0\geq\omega$, since $A$ is not a set.
Claim 2.2: For every set $c$ of atoms disjoint from $c_0$, there is a set $b$ of atoms disjoint from $c\cup c_0$, such that $b$ has cardinality $\mu_0$.
Proof: If not, let $c'=c\cup c_0$ and observe that $\mu_{c'}<\mu_0$, a contradiction.
From here we can run a simple variant of the proof from Case 1, but starting with $c_0\cup a_z$ as our base set of atoms. That is:
Claim 2.3: Let $a,a'$ be sets of atoms of the same cardinality, which are disjoint from $c_0$. Let $\pi:a\to a'$ a bijection, and $\sigma:V(c_0\cup a)\to V(c_0\cup a')$ the resulting isomorphism. Then for all formulas $\varphi$ and all $x\in V(c_0\cup a)$, we have $\varphi(x)\Leftrightarrow\varphi(\sigma(x))$ (with truth evaluated in the entire universe).
Proof: This is now proved like its version in Case 1, but using Claims 2.2 and 2.3 (the properties of $c_0$ and $\mu_0$) to see that we can find appropriate disjoint sets of atoms (of cardinality at most $\mu_0$).
The rest of the proof is like before; given a sequence $a_0,\ldots,a_n$ of sets of atoms disjoint from $c_0\cup a_z$, they each have cardinality $\leq\mu_0$, and we can always find another set $a$ of atoms of cardinality $\mu_0$ disjoint from the rest. (So in fact, since $\mu_0\geq\omega$, one could just fix a sequence $\left<a_i\right>_{n<\omega}$ of sets $a_i$ of atoms, each of cardinality $\mu_0$, disjoint from $c_0\cup a_z$ and pairwise disjoint, and then argue that given an ordinal $\beta$, there is an ordinal $\beta'$ such that for every $x\in V_{\beta}(c_0\cup a_z\cup a_0\cup\ldots\cup a_n)$ there is $y\in V_{\beta'}(c_0\cup a_z\cup a_0\cup\ldots\cup a_n\cup a_{n+1})$ such that $\varphi(x,y)$. Then we can define a resulting sequence $\left<\beta_n\right>_{n<\omega}$, and apply DC to sets as before.)
Remark: Considering @SamRoberts' and @EmilJeřábekand's comments below, we might have $\kappa_0=\mu_0=\omega$, hence no $\omega_1$-sequence of atoms. In this case the proof wouldn't work to give DC$_{\omega_1}$, since it would rely on having a sequence $\left<a_\beta\right>_{\beta<\omega_1}$ of pairwise disjoint (presumably non-empty) sets of atoms. We got a sequence like this above because we had the set of cardinality $\mu_0$ of atoms to work with, and since (in the argument) $\mu_0\geq\omega$, we can actually partition it as desired. But if $\mu_0=\omega$, this clearly breaks when we want an $\omega_1$-sequence.
