Yes. Let $S$ be our chosen foundational system. Even if we somehow know that $S$ is $Σ_1$-sound, we still cannot rule out the possibility that $S$ is $Σ_2$-unsound! Here truth and arithmetical soundness are of course defined with respect to the natural numbers in the meta-system.
First consider any ω-consistent formal system $T$ that has a proof verifier program and uniformly interprets PA. We shall construct an ω-consistent formal system $T'$ that is unsound.
For any sentence $P$ over $T$ we say that $T \vdash_ω P$ iff there is some $1$-parameter sentence $Q$ such that $T \vdash Q(0),Q(1),Q(2),\cdots$ and $∀k(Q(k)) \vdash P$, namely iff there is a proof of $P$ over $T$ using at most one application of the ω-rule. Now this itself can be expressed as a $Σ_3$-sentence, namely that there is a $1$-parameter arithmetic sentence $ωProv_T$ such that $T \vdash_ω P$ iff $\mathbb{N} \vDash ωProv_T(\ulcorner P \urcorner)$. We can write this succinctly using a new modal operator $ω⬜$, where $T \vdash_ω P$ iff $\mathbb{N} \vDash ω⬜_T P$.
Then it can be shown that $T \vdash_ω$ and $ω⬜_T$ satisfy the modal fixed-point theorem and the Hilbert-Bernays provability conditions, namely:
$
\def\imp{\Rightarrow}
$
If $T \vdash_ω P$ then $T \vdash_ω ω⬜_T P$.
$T \vdash_ω ω⬜_T P \land ω⬜_T ( P \imp Q ) \imp ω⬜_T Q$.
$T \vdash_ω ω⬜_T P \imp ω⬜_T ω⬜_T P$.
Thus the same proof of Lob's theorem immediately shows that if $T \vdash_ω ω⬜_T P \imp P$ then $T \vdash_ω P$. Since $T$ is ω-consistent, we have $T \nvdash_ω \bot$ and hence $T \nvdash_ω \neg ω⬜_T \bot$. Thus $T' \overset{def}= T + ω⬜_T \bot$ is ω-consistent. But by ω-consistency of $T$ we have $\mathbb{N} \nvDash ω⬜_T \bot$, and hence $T'$ is $Σ_3$-unsound.
Side remark: Since $\mathbb{N} \vDash \neg ω⬜_T \bot$, we also have that $T'' \overset{def}= T + \neg ω⬜_T \bot$ is ω-consistent, and so $T'$ and $T''$ are mutually incompatible ω-consistent extensions of $T$.
Next notice that you cannot possibly know whether your $S$ is not like $T'$, based on just knowing that $S$ is $Σ_1$-sound. So it is possible that $S$ is like $T'$ in being $Σ_3$-unsound and yet prove only true $Σ_1$-sentences. We can however generate all proofs of $S$ and see whether $S \vdash ω⬜_S \bot$. That would tell us that $S$ is unsound. If $S$ also interprets the weak predicative system ACA, we could generate all proofs and see whether $S$ proves itself arithmetically unsound, which would tell us that it has no $ω$-model.
A stronger but non-constructive approach is to note that there is no program using only the halting oracle $H$ that can determine the truth-value of any $Σ_2$-sentence, because that would contradict the unsolvability of the halting problem relative to $H$.
Now take any $Σ_1$-sound formal system $S$ that has a proof verifier program and interprets PA. Let $S'$ be $S$ plus all true $Π_1$-sentences. Then $S'$ is consistent, because otherwise $S$ proves some finite disjunction of false $Σ_1$-sentences, which is equivalent to a single false $Σ_1$-sentence. Also there is a program $V$ using $H$ that enumerates the theorems of $S'$, which includes all true $Σ_2$-sentences. By the consistency of $S'$ and the above fact, $V$ will fail to enumerate all the true $Π_2$-sentences. Thus there is a $Π_2$-sentence $U$ that is true but not provable in $S'$. Then $S+¬U$ is clearly not $Σ_2$-sound. Also, if $S+¬U$ is $Σ_1$-unsound, then $S+¬U$ proves some false $Σ_1$-sentence $F$, and hence $S+¬F$ proves $U$, which is impossible by choice of $U$. Thus $S+¬U$ is $Σ_1$-sound.
It turns out that this non-constructive approach can be made into a completely constructive one as described in this answer, which in short is by unfolding the constructive proof of the incompleteness theorem using the zero-guessing problem, relativized to the halting oracle, to construct an explicit true $Π_2$-sentence that $S'$ cannot prove. Hence the follow-up question in my comment is answered!
