Relative homology $H_n(T^2,S^1)$ (Hatcher, Exc 2.1,17.b)

Yes. Since $\mathbb Z$ is free, the sequence splits and you get what you want.

In case you're not aware of the splitting lemma, let's prove this by hand (this proof obviously extends easily to the fully general case).

First, let $H = \tilde{H}_1(T,S)$ and let's give the maps names:

$$0\rightarrow \mathbb{Z}\oplus\mathbb{Z} \overset{f_1}{\rightarrow} H \overset{f_2}{\rightarrow}\mathbb{Z}\rightarrow 0.$$

Note that the map $\mathbb Z \to 0$ is trivial so its kernel is the whole of $\mathbb{Z}$. Hence, the map $H \overset{f_2}{\to} \mathbb Z$ must be surjective by exactness. Let $h \in H$ be an element such that $f_2(h) = 1$. Define the function $g \colon \mathbb{Z} \to H$ on the generator by $g(1) = h$. It is then easy to see that $f_2(g(n)) = n$ for all $n \in \mathbb{Z}$.

Now define a map $i \colon (\mathbb{Z}\oplus \mathbb{Z}) \oplus \mathbb{Z} \to H$ by $i(x,y) = f_1(x) + g(y)$. We see that $i$ is clearly a homormophism.

Note that if $i(x,y) = 0$ then $f_1(x) + g(y) = 0$ which means that $f_1(x)=g(y)$ and so $f_2(f_1(x)) = f_2(g(y))$. But $f_2(g(y))=y$ by the definition of $g$. Also, by exactness, $f_2(f_1(x)) = 0$, so we have $y=0$. It follows that $f_1(x) = 0$ and, as $f_1$ is injective by exactness, $x$ must be $0$. It follows that $i$ is injective.

Now, let $z \in H$. Let $\tilde{z} = z - g(f_2(z))$. We have $f_2(\tilde{z}) = f_2(z) - f_2(g(f_2(z))) = f_2(z) - f_2(z) = 0$ so $\tilde{z} \in \ker f_2$. It follows that $\tilde{z} \in \operatorname{im}f_1$ by exactness. Let $a \in \mathbb{Z} \oplus \mathbb{Z}$ be such that $f_1(a) = \tilde{z}$. Finally, we see that $$i(a,f_2(z)) = f_1(a)+g(f_2(z)) = \tilde{z} + g(f_2(z)) = z - g(f_2(z)) +g(f_2(z)) = z.$$ Hence, $z$ is in the image of $i$ and so $i$ is surjective. It follows that $i$ is an isomorphism.