As Emil pointed out, assuming $\Sigma=\Sigma_1\cap\Sigma_2$, the union of 2 such theories is always conservative extension, Emil had proven it using the standard formulation Robinson’s joint consistency theorem, some places first prove the conservative extension variation and then uses it to get to the standard formulation, so here is a proof of this variation without assuming the Robinson’s joint consistency theorem:

Let $\phi$ be a sentence in $\Sigma$ provable in $T'$ and not provable in $T$.

Because $T_1$ is conservative, $\phi$ is also not provable there, but because $T_2 \cup (T_1\cup \{\lnot\phi\})$ is inconsistent there must be some $\psi$ provable in $T_2$ such that $\lnot\psi$ is provable in $T_1\cup\{\lnot\phi\}$ (this is a surprisingly tricky lemma when $T_1,T_2$ are not with the same language, it follows from Craig's interpolation), in particular $\psi$ and $\lnot\psi$ are sentences in $\Sigma$. Furthermore we have that $\lnot\phi\implies\lnot\psi$ is provable in $T_1$ and in the language $\Sigma$.

But because $T_2$ is conservative it means that $\psi$ is provable in $T$, and because $T_1$ is conservative we have $\lnot\phi\implies\lnot\psi$ also provable in $T$, but from $\psi$ and $\lnot\phi\implies\lnot\psi$ follows $\phi$ in $T$.

Note that the standard variation of the joint consistency theorem follows immediately, as any (consistent) extension of a complete theory is conservative.