Consider the two inclusions $\eta_i:\mathbb{S}^1\to \mathbb{S}^1\vee\mathbb{S}^1$. I claim that the following map is injective
$$(\eta_1\sqcup\eta_2)^*:[\mathbb{S}^1\vee\mathbb{S}^1,\mathbb{S}^1\vee\mathbb{S}^1]\to [\mathbb{S}^1\sqcup\mathbb{S}^1,\mathbb{S}^1\vee\mathbb{S}^1].$$
In other words, two self-maps $f,g:\mathbb{S}^1\vee\mathbb{S}^1\to \mathbb{S}^1\vee\mathbb{S}^1$ are homotopic if for the restrictions $f_i:=f\circ\eta_i$ and $g_i:=g\circ \eta_i$, we have $f_i\simeq g_i$.
If we would work with based homotopy, the statement would be clear. However, the homotopies $f_1\simeq g_1$ and $f_2\simeq g_2$ may move the base point …
Best Answer
It is not injective. I am reference Hatcher's section 4.A1 throughout which talks about basepointed vs. nonbasepointed homotopy classes of maps.
Let $f: S^1 \vee S^1 \rightarrow S^1 \vee S^1$ be $l \vee r$ where $l$ means wrap around the left copy of $S^1$ and $r$ the right, and $g:S^1 \vee S^1 \rightarrow S^1 \vee S^1$ be $l \vee (lr)^{-1}r(lr)$. The image of $[f]$ and $[g]$ under the map is the same since $[l]=[l]$ and $[(lr)^{-1}r(lr)]=[r]$ (since one is the conjugate of the other, see Hatcher).
Now $f$ is not homotopic to $g$ since if they were we could look at the path of the basepoint under such a homotopy. It suffices to show this path, really a loop, does not commute with $[l]:S^1 \rightarrow S^1 \vee S^1$, since then conjugation by this loop changes the basepointed homotopy type of $l$, meaning that there is no homotopy starting at $l$, ending at $l$, and taking the basepoint along such a path (see Hatcher for the relation between conjugation and homotopy).
A homotopy from $r$ to $(lr)^{-1}r(lr)$ necessarily has the homotopy class of the loop of the basepoint equal to a word with $r$ in it. This is because we are conjugating $r$ by a word to get $(lr)^{-1}r(lr)$ (again, see Hatcher for this relation). Such a thing necessarily does not commute with $l$, so there is no homotopy from $l \vee r$ to $l \vee (lr)^{-1}r(lr)$.