Under what conditions $\pi_1(X \vee Y) \approx \pi_1(X) * \pi_1(Y)$? 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?
 A: 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 $= \{y_0\} = \varphi^{-1}(0)$ 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.

*

*


*is automatically satisfied for metrizable spaces.



*

*

*is satisfied if there exists an open neighborhood $U$ of $y_0$ in $Y$ such that $(U,y_0)$ is pointed contractible. This condition also occurs in the Proposition of the question.



[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.
A: It is easy to see that while the assumptions of the statements in the original question and in Paul's answer are sufficient for the conclusion of the Seifert-van Kampen Theorem (SvK), they also give something far stronger. For example, both sets of assumptions imply that there are isomorphisms in homology
$$\tilde H_*(X\vee Y)\xrightarrow\cong\tilde H_*X\oplus\tilde H_*Y.$$
Generally, one might suspect that sufficient assumptions on the local properties of the wedge point for SvK to hold would only yield the above isomorphisms for $\ast=1$.
Now, the analogy between homology and homotopy can only be pushed so far. However, that the general assumptions seem stronger than necessary can be seen explicitly in Paul's answer, where they are strong enough to render the map $R$ a pointed homotopy equivalence when it is only required to induce an isomorphism on $\pi_1$.
Various attempts to weaken these assumptions have appeared in the literature. In particular, in line with the above reasoning, it was sought to replace the pointed contractible neighbourhoods of the basepoints with neighbourhoods which were only sufficiently homotopically trivial. The following partial generalisation is due to Griffiths [1], [2].

Theorem (Griffiths): Let $(X,x_0),(Y,y_0)$ be pointed spaces. Suppose that $X$ is both locally simply connected at $x_0$ and first-countable at $x_0$. Then the homomorphism
$$\phi:\pi_1(X,x_0)\ast\pi_1(Y,y_0)\rightarrow \pi_1((X,x_0)\vee(Y,y_0))$$
is an isomorphism. $\blacksquare$

Here, $X$ is said to be locally simply connected at $x_0$ if for any neighbourhood $U\subseteq X$ of $x_0$ there exists a neighbourhood $V$ of $x_0$ such that $x_0\in V\subseteq U$ and any loop in $V$ is homotopic in $U$ to a constant loop.
Any locally contractible space is locally simply connected. If $X$ is a metric space and the inclusion $\{x_0\}\subseteq X$ is a cofibration, then $X$ is locally simply connected at $x_0$. The Griffiths twin cone, as discussed above, is not locally simply connected at its basepoint. Indeed, it was Griffiths who introduced this space in [1] as a demonstration of the necessity of local simple connectedness in his theorem.
Now, although the theorem above includes no assumptions on $Y$ nor its basepoint $y_0$, there is one additional assumption on $X$ as compared to the previously discussed conditions. Namely, $X$ is required to be first-countable at its basepoint $x_0$. Of course, this holds trivially when $X$ is any metric space. However, it can fail even when $X$ is a nice, non-metrisable space such as a CW complex. In any case, we cannot view Griffiths's theorem as a true generalisation of the Seifert-van Kampen Theorem.
Griffiths suggested that the first-countability assumption was merely an artefact of his method of proof. It is perhaps surprising that this assumption is actually necessary in his formulation of the SvK Theorem.

Example (Eda [3]): There is a path-connected Tychonoff space $X$ and a point $x_0\in X$ such that $(i)$ $X$ is locally simply connected at $x_0$, $(ii)$ $\pi_1(X,x_0)=0$, and $(iii)$ $\pi_1((X,x_0)\vee(X,x_0))\neq0$. $\blacksquare$

[1] H. Griffiths, The fundamental group of two spaces with a common point, Quart. J. Math. Oxford 5 (1954), 175-190.
[2] H. Griffiths, The fundamental group of two spaces with a common point: a correction, Quart. J. Math. Oxford 6 (1955), 154-155.
[3] K. Eda, First countability and local simple connectivity of one point unions, Proc. Amer. Math. Soc. 109 (1990), 237-241.
A: The two previous answers to the question show that $\phi$ is an isomorphism provided one the following conditions is satisfied:
(1) $(Y,y_0)$ is well-pointed.
(2) $Y$ is both locally simply connected at $y_0$ and first-countable at $y_0$.
I was not aware of Griffiths's theorem based on (2), but let me discuss the relationship between (1) and (2).
First note that (1) and (2) do not impose any restriction on $(X,x_0)$ so that we get a sort of "asymmetric" theorem. In contrast the Proposition in the question is symmetric in $(X,x_0)$ and $(Y,y_0)$.
Observation 1.
If $(Y,y_0)$ is well-pointed, then $Y$ is locally pointed contractible at $y_0$ and in particular locally simply connected at $y_0$.
From Lemma 4 in Strøm's paper it is known that $(Y,y_0)$ is well-pointed if and only if there exists a map  $\varphi : Y \to I$ mit $y_0 \in \varphi^{-1}(0)$ and a homotopy $H : Y \times I \to Y$ such that

*

*$H(y,0) = y$ for all $y \in Y$.

*$H(y_0,t) = y_0$ for all $t \in I$.

*$H(y,t) = y_0$ for all $(y,t) \in Y \times I$ with $t > \varphi(y)$.

Now define $\Omega = \varphi^{-1}([0,1))$. This is an open neighborhood of $y_0$ in $Y$. The restriction $H' = H \mid_{\Omega \times I} : \Omega \times I \to Y$ is a pointed homotopy from the inclusion $(\Omega,y_0) \to (Y,y_0)$  to the constant pointed map.
Let $U$ be an open neighborhood of $y_0$ in $Y$. Since $\{y_0\} \times I \subset G^{-1}(U)$, we find an open neighborhood $V \subset \Omega$ of $y_0$ in $Y$ such that $V \times I \subset G^{-1}(U)$. Thus the restriction $G_{V,U} : V \times I \to U$ is a pointed homotopy from the inclusion $(V,y_0) \to (U,y_0)$  to the constant pointed map. This means that $Y$ is locally pointed contractible at $y_0$.
Observation 2.
Pointed $CW$-complexes $(Y,y_0)$ are well-pointed, but in general not  first-countable at $y_0$.
Thus Griffith's theorem does not cover all relevant cases. We can only say that if $(Y,y_0)$ is well-pointed and first-countable at $y_0$, then the isomorphy of $\phi$ is a consequence of Griffith's theorem.
Observation 3.
There are spaces $(Y,y_0)$ which are not well-pointed, but satisfy (2).
An example is the two-dimensional Hawaiian earring $Y$ (see Union of Spheres is simply connected). This is simply connected and by "self-similarity" it is easy to see that it is locally simply connected at the basepoint $y_0 = (0,0,0)$. But $(Y,y_0)$ is not well-pointed because it is not locally pointed contractible at $y_0$.
The above obvservations show that conditions (1) and (2) are independent and the theorems based on them have a different scope.
Griffith's Theorem 3 says that

$\phi$ is always injective.

This allows to prove
Theorem. Let $(X,x_0)$ and $(Y,y_0)$ be based spaces. Suppose that $y_0$ has an open neighborhood $\Omega$ in $Y$ such that the inclusion $(\Omega,y_0) \to (Y,y_0)$  is pointed homotopic to the constant pointed map. Then $\phi$ is an isomorphism.
Note that the above assumption covers all well-pointed $(Y,y_0)$, but does not assume that $Y$ is first countable at $y_0$. Therefore the theorem generalizes the result based on (1), but is still independent from Griffith's result based on (2).
Proof of Theorem. It remains to show that $\phi$ is surjective.
Let $u : I  \to (W,*) = (X,x_0) \vee (Y,y_0)$ be a loop based at $*$.  The set $U = u^{-1}(W \setminus X)$ is open in $I$ and does not contain the boundary points $0,1$. The connected components of $U$ are pairwise disjoint open intervals $J_\alpha, \alpha \in A$. Let $A' = \{ \alpha \in A \mid u(J_\alpha) \cap (Y \setminus \Omega) \ne \emptyset \}$. We claim that $A'$ is finite.
Assume that $A'$ is infinite. Then there exists a sequence of pairwise distinct $\alpha_i \in A'$, $i \in \mathbb N$. Pick $s_i \in J_{\alpha_i}$ such that $u(s_i) \in Y \setminus \Omega$. Note that by construction each $J_\alpha$ with $\alpha \in A$ contains at most one $s_i$. The sequence $(s_i)$ has cluster point $\sigma \in I$. Since $u$ is continuous,  $u(\sigma)$ is a cluster point of $(u(s_i))$. Since all $u(s_i)$ are contained in the closed set $Y \setminus \Omega$, we get $u(\sigma) \in Y \setminus \Omega \subset W \setminus X$.
Thus $\sigma \in U$. Let $\alpha^* \in A$ be the index such that $\sigma \in J_{\alpha^*}$. Since $J_{\alpha^*}$ is an open neighborhood of $\sigma$, infinitely many $s_i$ must be contained in $J_{\alpha^*}$. This is a contradiction.
So let $J_i = (a_i,b_i)$, $i=1,\ldots, n$, be the finitely many components of $U$ with image under $u$ intersecting $Y \setminus \Omega$. We have $u(a_i) = u(b_i) = *$. Note that the $K_i = [a_i,b_i]$ can intersect only in boundary points. Let $G : \Omega \times I \to Y$ be a pointed homotopy deforming $\Omega$ into $y_0$. We can extend it by the stationary homotopy on $X$ to a homotopy $G' : (X \cup \Omega) \times I \to W$. Let $I' = I \setminus \bigcup_{i=1}^n J_i$. Then $u(I') \subset X \cup \Omega$. Define
$$H' : I' \times I \to W, H'(s,t) = G'(u(s),t) .$$
We can extend it by the stationary homotopy on $K = \bigcup_{i=1}^n K_i$ to a homotopy $H : I \times I \to W$.
By construction $u$ is homotopic rel. $\{0,1\}$ to $v : I \to W, v(s) = H(s,1)$.
The set $I'$ is the finite union of closed intervals $J'_j = [a'_j,b'_j]$, $j = 1,\ldots, m$, which may be degenerate (i.e. $a'_j = b'_j$). Omitting such degenerate $J'_j$, we obtain a finite collection of closed subintervals $L_k = [c_k, d_k] \subset I$, $k=1,\ldots,r$ which cover $I$ and either have the form $J_i$ or $J'_j$. Two such $L_k$ can intersect only in common boudary points. All boundary points of the $L_k$ are mapped by $v$ to $*$. W.l.o.g. we may assume that $c_1 = 0, d_k = c_{k+1}, d_r = 1$.
The restrictions $v_k = v \mid_{L_k}$ are loops based at $*$ with images either in $X$ or in $Y$. Thus
$$[u] = [v] = [v_1]\ldots [v_r]$$
which is an element of $\pi_1(X,x_0) * \pi_1(Y,y_0)$.
