Let $(X,x_0)$ and $(Y,y_0)$ based spaces. Their wedge (= coproduct in the category of based spaces) is the based space
$$(W,*) = (X,x_0) \vee (Y,y_0) = ((X \sqcup Y)/{\sim}, *)$$
where $\sqcup$ denotes disjoint union and $\sim$ identifies $x_0 \in X$ with $y_0 \in Y$ and $*$ denotes the common equivalence class of $x_0, y_0$.
There are canonical embeddings $i_X : (X,x_0) \to (W,*)$, $i_Y : (Y,y_0) \to (W,*)$ and canonical retractions $r_X : (W,*) \to (X,x_0)$ and $r_Y : (W,*) \to (Y,y_0)$ which map all points of $Y$ resp. $X$ to $x_0$ resp. $y_0$.
Since $r_X \circ i_X =id$ and $r_Y \circ i_Y =id$ , we see that the induced
$$(i_X)_* : \pi_1(X,x_0) \to \pi_1(W,*), \\ (i_Y)_* : \pi_1(Y,y_0) \to \pi_1(W,*) .$$
are injective. Since the free product is the coproduct in the category of groups, they induce a canonical homomorphism
$$\phi : \pi_1(X,x_0) * \pi_1(Y,y_0) \to \pi_1(W,*) . \tag{1}$$
Explictly, for each reduced word $w = g_1h_1 \ldots g_nh_n \in \pi_1(X,x_0) * \pi_1(Y,y_0)$ with $g_i \in \pi_1(X,x_0), h_i \in \pi_1(Y,y_0)$ we have $\phi(w) = (i_X)_*(g_1)(i_Y)_*(h_1) \ldots (i_X)_*(g_n)(i_Y)_*(h_n)$.
Recalling that the fundamental group only depends on the path component of the basepoint, it suffices to consider $(1)$ for pathwise connected spaces $X, Y$.
One could naively believe that the theorem of Seifert – van Kampen shows that $\phi$ is an isomorphism. However, to apply the theorem, we need to find pathwise connected open subsets $U, V \subset W$ which are homotopy equivalent to $X$ and $Y$, cover $W$ and have a simply connected intersection containing $*$. In general this is impossible.
A nice example showing that the wedge of simply connected spaces is not always simply connected is the Griffiths twin cone. See also Union of simply connected spaces at a point not simply connected.
What conditions on $(X,x_0)$ and $(Y,y_0)$ assure that $\phi$ is an isomorphism?
A well-known answer follows immediately from the theorem of Seifert – van Kampen:
Proposition: If there are open neighborhoods of $x_0$ in $X$ and of $y_0$ in $Y$ which are pointed contractible to $x_0$ and $y_0$, then $\phi$ is an isomorphism.
Question:
Are there alternative assumptions assuring that $\phi$ is an isomorphism?
Best Answer
Here is an alternative result.
Theorem. Let $(X,x_0)$ and $(Y,y_0)$ be based spaces. If $(Y,y_0)$ is well-pointed (which means that the inclusion $\{y_0\} \to Y$ is a cofibration), then $\phi$ is an isomorphism.
Proof. The wedge $(A,a_0) \vee (B,b_0)$ has a canonical basepoint. If we take another basepoint $\xi \in (A \sqcup B)/{\sim}$, we get based space denoted by $$(A,a_0) \vee_\xi (B,b_0) = )(A \sqcup B)/{\sim,\xi)} .$$
Let $(Y',1) = (Y,y_0) \vee_1 (I,0)$, where $1 \in I = [0,1]$. The canonical retraction $r : (Y',1) \to (Y,y_0)$ is a free homotopy equivalence. Since both spaces are well-pointed (note that $(Y',1)$ is always well-pointed, even if $(Y,y_0)$ is not), $r$ is a pointed homotopy equivalence. See [1] Proposition 0.19.
Therefore the map $R = id \vee r: (X,x_0) \vee (Y',1) \to (X,x_0) \vee (Y,y_0)$ is a pointed homotopy equivalence and the diagram
$\require{AMScd}$ \begin{CD} \pi_1(X) * \pi_1(Y') @>{\phi}>> \pi_1(X \vee Y') \\ @V{id * r_*}VV @VV{R_*}V \\ \pi_1(X) * \pi_1(Y) @>>{\phi}> \pi_1(X \vee Y) \end{CD}
commutes. The vertical arrows are isomorphisms, thus it suffices to show that the upper horizontal arrow is an isomorphism.
Let $u : I \to I, u(t) = 2t$ for $t \le 1/2$, $u(t) = 1$ for $t \ge 1/2$. Define $(X',1/2) = (X,x_0) \vee_{1/2} ([1/2,1],1)$ and $(Y'',1/2) = (Y,y_0) \vee_{1/2} ([0,1/2 ],0)$. We have $(X,x_0) \vee (Y',1) = (X',1/2) \vee (Y'',1/2)$ as unbased spaces. The map $\bar u : (X',1/2) \vee (Y'',1/2) \to (X,x_0) \vee (Y',1)$ obtained by taking the identity on $X$ and $Y$ and $u$ on $I$ is a pointed map and a free homotopy equivalence. Also the restrictions $\bar u_X : (X',1/2) \to (X,x_0)$ and $\bar u_Y : (Y'',1/2) \to (Y',1)$ are pointed maps and free homotopy equivalences. Thus all these maps induce isomorphisms on fundamental groups (see [1] Proposition 1.18). The following diagram commutes: $\require{AMScd}$ \begin{CD} \pi_1(X') * \pi_1(Y'') @>{\phi}>> \pi_1(X' \vee Y'') \\ @V{(\bar u_X)_* * (\bar u_Y)_*}VV @VV{\bar u_*}V \\ \pi_1(X) * \pi_1(Y') @>>{\phi}> \pi_1(X \vee Y') \end{CD}
The vertical arrows are isomorphisms and the upper horizontal arrow is an isomorphism by the Proposition in the question.
Corollary. Let $(X,x_0)$ and $(Y,y_0)$ be based spaces. Assume
There exists an open neighborhood $U$ of $y_0$ in $Y$ such that the inclusion $j : (U,y_0) \to (Y,y_0)$ is pointed homotopic to the constant pointed map.
There exists a continuous $\varphi : Y \to I$ such that $ \{y_0\} = \varphi^{-1}(1)$ and $Y \setminus U \subset \varphi^{-1}(0)$.
Then $\phi$ is an isomorphism.
Proof. It is known that 1. + 2. imply that $(Y,y_0)$ is well-pointed. See [3] and [2] Exercise 1.E.6.
Remarks.
[1] Hatcher, Allen. Algebraic topology.
[2] Spanier, Edwin H. Algebraic topology. Springer Science & Business Media, 1989.
[3] Strøm, Arne. Note on cofibrations II. Mathematica Scandinavica 22.1 (1968): 130-142.