I will give an elementary proof of the problem using the fact, that $T^2$ is a topological group and that its universal cover is contractible. We will start with some constructions in homotopy theory of topological groups, which are required to understand the proof given below and added here for convenience.
First note, that $\mathbb R^2$ forms a topological group under addition: $(x,y) + (x',y') := (x+x',y+y')$ and that $\mathbb Z^2$ is a discrete normal subgroup thereof. We identify $T^2$ with the quotient of $\mathbb R^2$ by $\mathbb Z^2$, so that $T^2$ again becomes a topological group under addition. Moreover the quotient map $p: \mathbb R^2 \to T^2$ becomes a group homomorphism and is easily seen to be the universal covering projection ($\mathbb Z^2$ discrete subgroup $\Rightarrow$ $p$ is covering projection; $\mathbb R^2$ is contractible $\Rightarrow$ $p$ is universal).
Next, we observe, that for $[f],[g]\in [(X,\ast),(T^2,0)]$ the sum $[f]+[g] := [f+g]$ is well defined, turning $[(X,\ast),(T^2,0)]$ into a group. The same arguments show that $[(X,\ast),(\mathbb R^2,0)]$ is a group under point wise addition of representatives as well (as is $[(X,\ast),(G,1)]$ for any topological group $G$ with unit $1$), and that the map $p_\sharp : [(X,\ast),(\mathbb R^2,0)] \to [(X,\ast),(T^2,0)]$ given by $p_\sharp([f]) := [p \circ f]$ is a group homomorphism.
Now $\pi_1(T^2,0) = [(S^1,1), (T^2,0)]$ is a group in two ways, by means of composition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha \ast \beta]$ and by means of point wise addition of (representatives of) loops $[\alpha], [\beta] \mapsto [\alpha + \beta]$.
Both operations share the same unit, the (class of the) constant loop sending everything to $0 \in T^2$ and denoted simply by $0: (S^1,1) \to (T^2,0)$. We can also observe, that for any loops $\alpha, \beta, \gamma, \delta$ we have $(\alpha + \beta) \ast (\gamma + \delta) = (\alpha + \gamma) \ast (\beta + \delta)$. Therefore $$[\alpha]+[\beta] = ([\alpha] \ast [0]) + ([0] \ast [\beta]) = ([\alpha] + [0]) \ast ([0] + [\beta]) = [\alpha] \ast [\beta],$$
hence the two operations are in fact the same on $\pi_1(T_2,0)$. The same argument can be used to show the analogous statement for $\pi_1(\mathbb R^2,0)$ (or $\pi_1(G,1)$ for any topological group $G$ with unit $1$).
Now back to the problem:
Given two maps $\varphi, \psi: T^2 \to T^2$, such that for some point $x \in T^2$, we have $\varphi(x) = \psi(x) = x$ and $\pi_1(\varphi) = \pi_1(\psi): \pi_1(T^2,x) \to \pi_1(T^2,x)$, we want to show $\varphi \simeq \psi$, where the homotopy can be taken relative to $x$. Replacing $\varphi$ with $\xi \mapsto \varphi(\xi + x) - x$ and $\psi$ with $\xi \mapsto \psi(\xi + x) - x$ if necessary, we may assume $x=0$. It will then suffice to show, that $\chi \simeq 0$, where $\chi := \varphi - \psi$.
Since the induced map $\pi_1(\chi): \pi_1(T^2,0) \to \pi_1(T^2,0)$ on fundamental groups is trivial (this is where we need all the constructions for topological groups), we can lift $\chi$ to a map $\bar{\chi}: (T^2,0) \to (\mathbb R^2,0)$ with $\chi = p \circ \bar{\chi}$.
We now define $H: T^2 \times I \to T^2$ by $H(x,t) = p(t\bar\chi(x))$, which is easily checked to be the required homotopy $0 \simeq \chi$.
Best Answer
You can prove this more general result:
Lemma. Let $A\subset X,$ $B\subset Y.$ If $f_0,f_1:X\to Y$ and $g_0,g_1:Y\to Z$ are maps, $f_0\simeq f_1$ rel $A,$ $g_0\simeq g_1$ rel $B,$ and $f_0(A)=f_1(A)\subset B,$ then $g_0\circ f_0\simeq g_1\circ f_1$ rel $A.$
Proof sketch. If $h:X\times I\to Y$ is a homotopy from $f_0$ to $f_1$ rel $A$ and $h':Y\times I\to Z$ is a homotopy from $g_0$ to $g_1$ rel $B,$ check that $H:X\times I\to Z$ given by $$H(x,t)=h'(h(x,t),t)$$ gives the desired homotopy from $g_0\circ f_0$ to $g_1\circ f_1$ rel $A.$