Let $a = S_a(0)$ and $b = S_b(0)$. We have $S_a(x) = x+a$ and $S_b(x) = x+b$ from (3) and (4), so $+$ is really concatenation and $a,b$ are the distinct symbols by (6).
Let (5') be the weaker version of (5). As you probably know, (5') is a trivial consequence of (5).
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With (5), your theory $T$ directly interprets TC (using the equivalent axiomatization here) because we can easily prove:
$\color{blue}{ ∀x\ ( \ 0+x = x \ ) }$.
Proof: Clearly we have $0+0 = 0$. Now take any $x$ such that $0+x = x$. Then we have $0+(x+a) = (0+x)+a = x+a$ and similarly $0+(x+b) = x+b$. Therefore by (5) we are done. $\small_\strut$
$\color{blue}{ ∀x,y,z\ ( \ x+(y+z) = (x+y)+z \ ) }$.
Proof: Clearly we have $∀x,y\ ( \ x+(y+0) = (x+y)+0 \ )$. Now take any $x,y,z$ such that $x+(y+z) = (x+y)+z$. Then $x+(y+(z+a)) = x+((y+z)+a)$ $= (x+(y+z))+a = ((x+y)+z)+a$ $= (x+y)+(z+a)$ and similarly $x+(y+(z+b)) = (x+y)+(z+b)$. Therefore by (5) we are done. $\small_\strut$
$\color{blue}{ ∀x,y\ ( \ a = x+y ∨ b = x+y ⇒ x = 0 ∨ y = 0 \ ) }$.
Proof: Take any $x,y$ such that $a = x+y$. If $y ≠ 0$ then $y = t+a$ or $y = t+b$ for some $t$ by (5'), and so $a = (x+t)+a$ or $a = (x+t)+b$, but the latter is impossible by (6), so $0 = x+t$ by (1). Thus $t = 0$ otherwise by (5') and (1) we get a contradiction. Thus $x = 0$. Similarly for any $x,y$ such that $b = x+y$. $\small_\strut$
$\color{blue}{ ∀x,y,z,w\ ( \ x+y = z+w ⇒ ∃c\ ( \ Q(x,y,z,w,c) \ ) \ ) }$ where $Q(x,y,z,w,c)$ $≡ x+c = z ∧ y = c+w ∨ x = z+c ∧ c+y = w$.
Proof: Clearly $∀x,y,z\ ( \ x+y = z+0 ⇒ Q(x,y,z,0,y) \ )$. Now take any $w$ such that $∀x,y,z\ ( \ x+y = z+w ⇒ ∃c\ ( \ Q(x,y,z,w,c) \ ) \ )$. And take any $x,y,z$ such that $x+y = z+(w+r)$ where $r = a$ or $r = b$. If $y = 0$, then $Q(x,y,z,w+r,w+r)$. If $y ≠ 0$, then $y = t+s$ for some $t,s$ such that $s = a$ or $s = b$ by (5'), and so we have $(x+t)+s = (z+w)+r$ and hence $s = r$ by (6) and $x+t = z+w$ by (1). Let $d$ be such that $Q(x,t,z,w,d)$. Then $Q(x,y,z,w+r,d)$. In either case, we have shown that $∃c\ ( \ Q(x,y,z,w+r,c) \ )$. Therefore by (5) we are done. $\small_\strut$
Since TC can reason about programs (as defined and sketched here), so can $T$, and hence $T$ is essentially undecidable (if consistent).
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If (5) is weakened to (5'), it is no longer clear that the resulting theory $T'$ interprets TC. But $T'$ can still reason about programs, for the same essential reason as TC. The substring relation $⊆$ is definable via $x ⊆ y ≡ ∃t,u\ ( \ (t+x)+u = y \ )$, and we shall write "$∀x{⊆}y\ ( \ Q(x) \ )$" to mean "$∀x\ ( \ x⊆y ⇒ Q(x) \ )$" and "$∃x{⊆}y\ ( \ Q(x) \ )$" to mean "$∃x\ ( \ x⊆y ∧ Q(x) \ )$", and we shall call "$∀x{⊆}y$" and "$∃x{⊆}y$" bounded quantifiers. We can use the program execution encoding sketched in the linked thread, by which we can for any given strings $p,x,y$ translate "The program $p$ halts on input $x$ and outputs $y$." to a $Σ_1$-sentence over $T'$, namely of the form "$∃h\ ( \ P(h) \ )$" where $P$ has only bounded quantifiers. (The unbounded quantifier "$∃h$" here is for the finite sequence of program states witnessing the entire history of the execution of $p$ on $x$.) Similarly we can translate "The program $p$ halts on input $x$." and "It is not true that the program $p$ halts on input $x$ and outputs $y$." to $Σ_1$-sentences over $T'$.
So all we need to show is that $T'$ can prove every true $Σ_1$-sentence over $T'$, where truth is with respect to the standard model (i.e. the structure of finite strings from the alphabet $\{a,b\}$). Since such a sentence has a constant term witness, it clearly suffices to show that $T'$ can prove every true sentence over $T'$ with only bounded quantifiers. This can be done by converting it to prenex normal form $E$ (while keeping the bounded quantifiers) with matrix in disjunctive normal form, and then inducting on the number of quantifiers.
If $E$ has no quantifiers, all the terms are constants and it is easy to check that $T'$ proves $E$ using (1),(2),(3),(4),(6) for inequalities.
Note that if $E$ has quantifiers, the bound for the first quantifier must be a constant term.
If $E$ is $∃x{⊆}m\ ( \ P(x) \ )$ for some constant term $m$ and predicate $P$ over $T'$, then there is a term witness $k$, so $T'$ proves $k ⊆ m$ (as it is witnessed by just an equality) and also proves $P(k)$ (since it is true and has fewer quantifiers than $E$), and hence proves $E$.
If $E$ is $∀x{⊆}m\ ( \ P(x) \ )$ for some constant term $m$ and predicate $P$ over $T'$, then we need the key fact that $T'$ proves $∀x{⊆}m\ ( \ \bigvee_{k{⊆}ι(m)} x = τ(k) \ )$, where $ι(m)$ is the interpretation of $m$ and $τ(k)$ is the standard term representing string $k$ (i.e. built from $0,S_a,S_b$). Given that fact, $T'$ proves $E$ because $T'$ proves $P(τ(k))$ for every $k⊆ι(m)$ (since $P(τ(k))$ is true and has fewer quantifiers than $E$).
The key fact can be proven via induction over the length of $m$, and this is where we finally need (5'). Note that we can use (3),(4) to prove equality between $m$ and its standard form, so we can assume that $m$ is a standard term. Take any $x ⊆ m$. Let $t,u$ be such that $(t+x)+u = m$. If $m$ is $0$, then we are done since $(t+x)+u = 0$ implies $u = 0$ by (5'),(4),(2), which implies $t+x = 0$ by (3) and hence similarly $x = 0$. So we can assume $m$ is not $0$. Then $m$ must be $S_s(n)$ for some term $n$ and $s∈\{a,b\}$. If $u = 0$, then $t+x = n+s$ by (3), so $x = v+s$ for some $v$ by (5'),(4),(6), and hence $t+v = n$ by (4),(1), yielding $v ⊆ n$ by (3). If $u ≠ 0$, then $u = w+s$ for some $w$ by (5'),(4),(6), and hence $(t+x)+w = n$ by (4),(1), yielding $x ⊆ n$. In either case, we can invoke the induction since $n$ is shorter than $m$.
