Let $\{Y_i\}_i$ be the collection of path components of $Y$, so then $X*Y=\bigcup_i X*Y_i$. Let $Z$ be the portion of $X*Y$ associated with $[0,1/2)$, and let $A_i=Z\cup (X*Y_i)$. The intersection of any two of these $A_i$ is $Z$, which is path connected (deformation retracting onto $X$), so if we prove that $X*Y_i$ is simply connected, then $X*Y$ is simply connected by the van Kampen theorem. While it is true that the $A_i$ are not open in $X*Y$, this is fine because the surjectivity part of the theorem only relies on $f^{-1}(A_i)$ being open in $[0,1]$ for any path $f:[0,1]\to X*Y$. We defer showing openness to the appendix.
Because of this, let us assume $Y$ is path connected as well. The space $X*Y$ decomposes into two open subsets, one associated with $[0,2/3)$ and the other associated with $(1/3,1]$. Call these $Z_1$ and $Z_2$. The first deformation retracts onto $X$, the second deformation retracts onto $Y$, and their intersection $Z_1\cap Z_2$ deformation retracts onto $X\times Y$. The inclusion maps $X\times Y\to Z_1$ and $X\times Y\to Z_2$ induce projections on $\pi_1$, so the van Kampen theorem gives $\pi_1(X*Y)$ to be the free product $\pi_1(X)*\pi_1(Y)$ modulo the subgroup generated by all $i_{1*}(x,y)i_{2*}(x,y)^{-1}$, which are all $xy^{-1}$. Since $x,y$ may be arbitrary, the resulting $\pi_1(X*Y)$ is trivial, so $X*Y$ is simply connected.
Appendix. First, $Z$ is open in $X*Y$, so $f^{-1}(Z)$ is open. Take a point $f(s)=(x,y,t)$ in $X*Y_i$ with $t>0$. By continuity of $f$ there is a $\delta$ such that $f((s-\delta,s+\delta))\subset X\times Y\times (0,1]$. Since the image of this interval must lie in a path component, $f((s-\delta,s+\delta))\subset A_i$.
This solution relies on the long exact reduced homology sequence of a NDR pair (Hatcher's Theorem $2.13$), and provides different approach to the problem from @tsho's solution.
Let us call $$\underbrace{\Huge{\mathsf x} \normalsize\times S^1}_{A}~~\subset~~
\underbrace{\Huge{\propto}\normalsize\times S^1}_{B}~~\subset~~
\underbrace{\Huge{\infty}\normalsize \times S^1}_{X}$$ All three pairs $(B,A),~(X,A),~(X,B)$ are Neighborhood Deformation Retracts, as Hatcher puts it, "good pairs". Also, it is obvious that $A$ is homotopy equivalent to the circle, and $B$ to the torus. Let us write the long exact reduced homology sequence for the the good pairs $(X,A)$ and $(X,B)$ : the morphism of pair given by the inclusion $(X,A)\hookrightarrow (X,B)$ gives following commutative diagram
$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/A)&\to&\tilde{H}_1(A)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/A)&\to 0\\
&\downarrow&&\Vert&&\downarrow&&\downarrow&&\Vert&&\downarrow\\
0\to& \tilde{H}_2(B)&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/B)&\to&\tilde{H}_1(B)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/B)&\to 0
\end{array}$$
Now $X/A$ is the wedge sum of two pinched spheres $P$ (the space studied in the previous question), and $X/B$ is simply a pinched sphere, and it follows that $\tilde{H}_*(X/B)\simeq\tilde H_*(P)$ and $\tilde{H}_*(X/A)\simeq\tilde H_*(P)\bigoplus \tilde H_*(P)$ where the isomorphism is given by the map $(i_*^+,i_*^-)$ where $i^+$ (resp. $i^-$) are the inclusions of $P$ as the upper (resp. lower) pinched sphere in $X/A$.
Since a pinched sphere is homotopy equivalent to a sphere with a diameter attached to it, which in turn is homotopy equivalent to the wedge sum of a sphere and a circle, we have $\tilde H_*(P)\simeq \Bbb Z\oplus\Bbb Z$ concentrated in degree $1$ and $2$. We can now replace the above diagram by the following simpler one
$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\Bbb Z\oplus \Bbb Z &\stackrel{\gamma}{\to}&\Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z\oplus \Bbb Z &\to 0\\
&\downarrow&&\Vert&&~~\downarrow\beta&&\downarrow&&\Vert&&\downarrow\\
0\to& \Bbb Z &\to&\tilde{H}_2(X)&\stackrel{\alpha}{\to}& \Bbb Z &\to& \Bbb Z\oplus \Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z &\to 0
\end{array}$$
From the left side of this diagram, it follows that $\tilde H_2(X)$ is a subgroup of $\Bbb Z\oplus \Bbb Z$ containing a copy of $\Bbb Z$, so $\tilde H_2(X)\simeq\Bbb Z$ or $\Bbb Z\oplus\Bbb Z$. Let us assume $\tilde H_2(X)\simeq \Bbb Z$ and try to derive a contradiction.
Since $\Bbb Z$ is torsion free, we must have $\alpha=0$. The vertical map $\beta$ is onto as it corresponds to collapsing the lower copy of $P$ inside $P\vee P\simeq X/A$ to a point, and thus $\beta$ is the projection onto the first factor. The commutativity of the diagram then forces the image of $\tilde H_2(X)$ to lie inside $\Bbb Z\oplus 0\subset \Bbb Z\oplus \Bbb Z$. However, there is an obvious self-homeomorphism of $X$ interchanging the upper and lower toruses of $X$ which passes to the quotient, and permutes the two factors $\Bbb Z \oplus \Bbb Z=\tilde H_2(X/A)$ (and possibly adds a sign). Thus, the image of $\tilde H_2(X)$ inside $\Bbb Z \oplus \Bbb Z$ is contained in $\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$, but this contradicts the injectivity of the top left arrow. The same argument works when we replace $B$ with $B'=T(B)$ where $T$ is the self-homeomorphism of $X$ that interchanges the two circles in the wedge sum $S^1\vee S^1$. The new map $\beta'$ is the projection onto the second factor, so the map
$\tilde H_2(X)\to \Bbb Z\oplus\Bbb Z$ sends $\tilde H_2(X)$ into $\ker(\beta)\cap\ker(\beta')=\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$ contradicting injectivity.
As a consequence, $$\tilde H_2(X)\simeq \Bbb Z\oplus\Bbb Z$$
To finish the proof, we note that by the standard theory of finitely generated abelian groups, the quotient of $\Bbb Z\oplus\Bbb Z$ by a subgroup $S$ isomorphic to $\Bbb Z\oplus\Bbb Z$ is the product of two cyclic groups, and cannot be a subgroup of $\Bbb Z$ unless the subgroup $S$ is all of $\Bbb Z\oplus\Bbb Z$. This forces the top left arrow $\tilde{H}_2(X)\to\Bbb Z\oplus \Bbb Z $ to be an isomorphism, and $\gamma=0$. The top sequence then degenerates to a short exact sequence
$$0\to\Bbb Z \to\tilde{H}_1(X)\to \Bbb Z\oplus \Bbb Z \to 0$$
Consequently, $$\tilde{H}_1(X)\simeq \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$$
Best Answer
The freakish argument in the comments is more elementary, but I thought you might appreciate an argument using a Mayer-Vietoris sequence.
Let $n=\dim X$, let $\varphi:U\to\mathbb R^n$ be a chart at $x$ with $\varphi(x)=0$, and let $V=X\setminus\varphi^{-1}(\overline B(0,1))$. Then $U\approx\mathbb R^n$, $V\approx\tilde X$, $U\cap V\approx S^{n-1}$. Since $\pi_1X=0$, the Hurewicz theorem implies $H_1(X)=0$, so the Mayer-Vietoris sequence for the decomposition $X=U\cup V$ gives us a short exact sequence $$0\to H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0.$$ With the above identifications, this gives us $$0\to\mathbb Z\to\mathbb Z\oplus H_0(\tilde X)\to\mathbb Z\to 0,$$ forcing $H_0(\tilde X)$ to be $\mathbb Z$.