The original version of this answer contained a mistake, as pointed out by Hanul Jeon.
The main issue was that I was assuming too much of the $\omega$ of the model. This version makes a weaker claim.
I think by what you write, that by "$\Gamma$-Foundation" you mean that for each $\Gamma$ formula $\varphi$, we have the axiom "for all $p$, if
$$\forall x\ \Big[(\forall y\in x\ \varphi(p,y))\Rightarrow\varphi(p,x)\Big],$$
then $$\forall x\ \varphi(p,x).\text{"}$$
I will anyway call scheme "$\Gamma$-Induction".
Claim: (i) $\mathrm{KP}_{\omega_0}+\Pi_1$-Induction does not prove $\Sigma_1$-Induction.
However, (ii) the consistency strength of $\mathrm{KP}_{\omega_0}+\Sigma_1$-Induction is at least
as strong as $\mathrm{KP}_{\omega_0}+\Sigma_1$-Induction + $\Pi_1$-Induction,
as every model of the former
has an inner model of the latter (its constructible universe $L$,
defined in a natural manner).
Here "$\Gamma$-Induction" is the scheme where for each $\Gamma$ formula $\varphi$, we have the axiom "for all $p$, if
$$ \forall x\ \Big[(\forall y\in x\ \varphi(p,y))\Rightarrow\varphi(p,x)\Big],$$
then $$ \forall x\ \varphi(p,x).\text{"}.$$
I also want to formulate a variant of the main claim.
Working in KP$_{\omega_0}$,
define (von Neumann) ordinals
as usual (transitive sets which are linearly ordered by $\in$).
Lemma 1: KP$_{\omega_0}$ proves
that ordinals are comparable (via $\in$), and that there is a least limit ordinal, which we denote $\omega$.
Proof: This is as usual, noting that KP$_{\omega_0}$ suffices.
Here it is, for convenience: Given ordinals $\alpha,\beta$, let $\gamma=\alpha\cap\beta$,
and note that $\gamma$ is an ordinal, but $\gamma\notin\alpha\cap\beta$.
We must have $\gamma=\alpha$
or $\gamma=\beta$. For otherwise,
by Foundation we can let $\delta\in\alpha\backslash\gamma$ be $\in$-minimal
and $\varepsilon\in\beta\backslash\gamma$ be $\in$-minimal.
We can't have $\delta=\varepsilon$
(by definition of $\gamma$),
but we do have $\gamma\subseteq\delta\cap\varepsilon$. But this easily leads to a contradiction to the $\in$-minimalities of these two. So either $\gamma=\alpha$
or $\gamma=\beta$.
Say $\gamma=\alpha$.
We may assume $\gamma\neq\beta$.
Now argue similarly as before
to see that $\gamma=\alpha\in\beta$.
Now by Infinity,
there is an inductive set,
i.e. a set $w$ with $0\in w$
and $w$ closed under $+1$.
Let $w'$ be the set of ordinals
in $w$; note that this is also inductive. Let $w''$
be the set of all $x\in w'$
such that $x\subseteq w'$;
note that this is also inductive.
Note that $w''$ is a transitive
set of ordinals, so $w''$ is an ordinal (using the comparability of ordinals). Note that $w''$ is a limit ordinal. If $w''$ contains no limit ordinals, then it is the least limit ordinal, by comparability. Otherwise, let $X$ be the set of limit ordinals in $w''$. Then by Foundation, there is an $\in$-minimal element $\eta$ of $X$. This is the least limit ordinal.
Now $\Gamma$-Induction over $\omega$ is the scheme where for each $\Gamma$-formula $\varphi$,
we have the axiom "for all $p$, if
$$ \varphi(p,0)\wedge \forall n\in\omega\ \Big[\varphi(p,n)\Rightarrow\varphi(p,n+1)\Big],$$
then $$ \forall n\in\omega\ \varphi(p,n).\text{"}.$$
Lemma 2: KP$_{\omega_0}$ + $\Sigma_1$-Induction proves $\Sigma_1$-Induction over $\omega$.
Proof: Let $\varphi$
be $\Sigma_1$ and $p$ be a set
and suppose that
$$ \varphi(p,0) $$
and
for all $n\in\omega$,
$$\text{ if }\varphi(p,n)\text{ then }\varphi(p,n+1).$$
Let $\varphi'$ be the formula
in parameters $p,\omega$ asserting
``if $x\in\omega$
then $\varphi(p,x)$''.
Note that for all sets $x$,
if $\varphi'(p,y)$ holds for all $y\in x$, then $\varphi'(p,x)$ holds. So we can apply $\Sigma_1$-Induction, giving that $\varphi'(p,x)$ holds for all $x$,
which clearly suffices.
Claim': (i) KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$ + $\Pi_1$-Induction does not prove $\Sigma_1$-Induction.
(ii) KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$
has consistency strength at least as strong as KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$ + $\Pi_1$-Induction. In fact every model of the former has an inner model satisfying the latter (its $L$).
We first prove (ii) of the Claim and Claim'.
Lemma 3: KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$ proves
that transitive closures exist.
Proof: Let $x$ be any set.
Define by recursion $\left<x_n\right>_{n<\omega}$,
where $x_0=x$ and $x_{n+1}=x_n\cup \bigcup x_n$. This runs all the way through $\omega$, by $\Sigma_1$-Induction over $\omega$. Now apply KP to the fact that for all $n\in\omega$ there is a sequence $\left<x_i\right>_{i\leq n}$ satisfying the above recursion through to $n$,
to get that there is a set $X$ such that for all $n<\omega$,
we have $x_n\in X$ and $\left<x_i\right>_{i\leq n}x_n\in X$. Now just use $\Delta_0$-Separation with the set $X\cup\bigcup X$
to see that the transitive closure of $x$ exists (it is a subset of $\bigcup X$).\
Now I want to sensibly define $L^M$ for a model $M$ of KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$.
Given $\alpha\in\mathrm{OR}^M$,
say $\alpha$ is good if there is a sequence $\left<l_\beta\right>_{\beta\leq\alpha}$ satisfying the usual recursion for the levels of $L$.
If $\alpha$ is good, then there is a unique witnessing sequence.
So for good $\alpha$, define $L_\alpha^M$ as the unique witnessing $l_\alpha$. Now let $G$ be the class of all good $\alpha$ of $M$, and define $L^M=\bigcup_{\alpha\in G}L_\alpha^M$. Note that $G$
is closed downward by definition,
and by
$\Sigma_1$-Induction
over $\omega$, it has no largest element (for each $L_\alpha^M$, the satisfaction relation for $L_\alpha^M$ exists as a set, so we can define $L_{\alpha+1}^M$).
Lemma 4: $\mathrm{KP}_{\omega_0}$+ $\Sigma_1$-Induction over $\omega$ + "$V=L$" proves $\Pi_1$-Induction.
Proof: Let $M$ be a model of $\mathrm{KP}_{\omega_0}$+ $\Sigma_1$-Induction over $\omega$ + "$V=L$".
Let $\varphi$ be $\Pi_1$ and $p\in M$ and suppose that $\forall x\in M$,
if $M\models\forall y\in x\ \varphi(y,p)$, then $M\models\varphi(x,p)$.
Suppose there is some $x\in M$ such that $M\models\neg\varphi(x,p)$,
and fix such an $x_0$.
Let $X_0=\{x_0\}$.
Working in $M$, define a sequence $\left<X_n\right>_{n<\omega}$ of non-empty sets $X_n$ as follows.
We will maintain by recursion that $X_n\in M$ and $M\models$ "$\forall x\in X_n\ \neg \varphi(x,p)$". Clearly we have this at $n=0$ already. So, suppose we have $X_n$. Let $\psi(x,p)$ be the formula "$\exists y\in x\ [\neg\varphi(y,p)]$", which is $\Sigma_1$. Note that by KP + "$V=L$",
there is $\eta<\mathrm{OR}^M$ such that $M\models$ "$\forall x\in X_n\ \Big[L_\eta\models\psi(x,p)\Big]$",
so $M\models$ "$\forall x\in X_n\ \exists y\in x\ \Big[L_\eta\models\neg\varphi(y,p)\Big]$".
Let $\eta_n$ be the least such $\eta$ with also $\eta>\eta_{n-1}$ if $n>0$. Now let $X_{n+1}$ be the set of all $y$ such that for some $x\in X_n$, we have $y\in x$
and $L_{\eta_n}\models$"$\neg\varphi(y,p)$". By choice, $X_{n+1}\neq\emptyset$
and for all $y\in X_{n+1}$,
since $L_{\eta_n}\models$"$\neg\varphi(y,p)$" and $\neg\varphi$ is $\Sigma_1$,
also $M\models$"$\neg\varphi(y,p)$". This completes the recursion.
Since $M\models$ KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$, and the whole construction is a $\Sigma_1$-recursion of length $\omega$, we get that $\sup_{n<\omega}\eta_n<\mathrm{OR}^M$ and $\left<X_n\right>_{n<\omega}\in M$.
But now let $X=\bigcup_{n<\omega}X_n$ and note that $X$ violates Foundation. It is, moreover,
easy to see that Foundation follows from $\Delta_0$-Induction, which is included in the axioms.
Lemma 5: Let $M$
be a model of KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$.
Then $L^M\models$ KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$ + "$V=L$", and hence (by Lemma 4) $L^M$ also models $\Pi_1$-Induction. Moreover,
if $M\models\Sigma_1$-Induction
then every ordinal of $M$
is good, and $L^M\models\Sigma_1$-Induction.
Proof: The main three things to verify are (a) $L^M\models$ KP$_{\omega_0}$, (b)
$L^M\models$ Infinity,
and (c) $L^M\models \Sigma_1$-Induction (possibly just over $\omega$).
We first consider (a).
For this, the main issue is $\Sigma_1$-bounding.
Sublemma:
There is no $x\in M$ such that for cofinally many good $\alpha\in G$, we have $L_\alpha^M\in x$.
Proof: Suppose otherwise.
Then there is there is a set $x\in M$ such that $L_\alpha^M\in x$
for cofinally many good $\alpha$.
Let $x'$ be the transitive closure of $x$, so then $L^M\subseteq x'$.
In particular, every good $\alpha$ is in $L^M$.
And note that the class $G$ of all good $\alpha$ is definable over $x'$, since the recursions needed to witness their goodness are all in $L^M$, so in $x'$. So $G\in M$.
But then $G\in\mathrm{OR}^M$
and again using the properties of $x'$, it is easy to see that $G$ is also good, a contradiction.
This proves the sublemma.
We can now verify $\Sigma_1$-bounding in $L^M$.
So let $\varphi$ be $\Sigma_1$
and $d,p\in L^M$
and suppose (i) $L^M\models$ "For all $x\in d$
there is $y$ such that $\varphi(p,x,y)$". We must see that (ii) there is $w\in L^M$ such that $L^M\models$" For all $x\in d$
there is $y\in w$ such that $\varphi(p,x,y)$".
So assume (i) holds.
Now $L^M$ is $\Sigma_1^M$,
as is the class of good $\alpha$
and the function $\alpha\mapsto L_\alpha^M$ for good $\alpha$.
So note that $M\models$"For all $x\in d$ there is a good $\alpha$
such that $L_\alpha^M\models\varphi(p,x,y)$". So by KP in $M$, there is a set $w$ such that $M\models$"For all $x\in d$ there is a good $\alpha\in w$ such that $L_\alpha^M\models\varphi(p,x,y)$". But then by the sublemma,
the set $w$ can only contain boundedly many good $\alpha$s,
so letting $\beta$ be good
and containing all of those,
we get that $L_\beta^M\models$"For all $x\in d$ there is $y$ such that $\varphi(p,x,y)$", which suffices.
Now consider (b),
i.e. the Axiom of Infinity.
For this, it is enough to see that $\omega\in L^M$.
But $\omega\subseteq G$
by $\Sigma_1$-Induction over $\omega$ in $M$, and so $\omega\subseteq L^M$,
and then clearly also $\omega\in G$, so $\omega\in L^M$.
Now we verify that $L^M\models\Sigma_1$-Induction over $\omega$. But this holds because $\omega^{L^M}=\omega$, and since $L^M$ is transitive
and $\Sigma_1^M$-definable.
Finally suppose that $M\models\Sigma_1$-Induction.
Then clearly $G=\mathrm{OR}^M$
and the fact that $L^M\models\Sigma_1$-Induction is like in the previous paragraph.
For (c): That $L^M\models$ $\Sigma_1$-Induction over $\omega$
is a straightforward
relativization,
using that $L^M$ is $\Sigma_1^M$-definable.
For full $\Sigma_1$-Induction,
suppose $M\models\Sigma_1$-Induction, and let $\varphi$ be $\Sigma_1$ and $p\in L^M$,
and suppose that $L^M\models$"For all $x$, if for all $y\in x$, $\varphi(p,y)$ holds, then $\varphi(p,x)$ holds".
Now suppose for contradiction
that $x_0\in L^M$ but $L^M\models\neg\varphi(p,x_0)$.
Let $\alpha$ be good and large enough that $p,x_0\in L_\alpha^M$.
Let $\varphi'(p,x,L_\alpha^M)$
assert (of parameters $p,L_\alpha^M$
and variable $x$)
"if $x\in L_\alpha^M$
then $L\models\varphi(p,x)$".
Then note that $M\models$ "for all $x$, if for all $y\in x$, $\varphi'(p,y)$ holds,
then $\varphi'(p,x)$ holds".
(For let $x\in L_\alpha^M$
and suppose that $M\models\varphi'(p,y)$ for all $y\in x$. Then since $x\in L_\alpha^M$,
we have $x\subseteq L_\alpha^M$,
and therefore $L^M\models\varphi(p,y)$ for all $y\in x$, and so $L^M\models\varphi(p,x)$, which suffices.) Therefore by $\Sigma_1$-Induction in $M$,
$M\models$ "For all sets $x$,
$\varphi'(p,x)$ holds",
so $L^M\models$ "For all sets $x\in L_\alpha^M$, $\varphi(p,x)$ holds", contradicting
that $x_0\in L_\alpha^M$.
This completes the proof of Lemma 5, and therefore of Claim (ii) and Claim' (ii).
Claim (i) is a direct consequencde of Claim' (i). So it just remains to see:
Proof of Claim' (i): Let $M$
be any model of KP + "$V=L$" which is illfounded and having wellfounded part $L_{\omega_1^{\mathrm{ck}}}$.
Let $\eta$ be an illfounded ordinal of $M$. Let $N=L_{\eta+\omega_1^{\mathrm{ck}}}^M$
(that is, $N$ is the "union"
of all segments $L_{\eta+\alpha}^M$, for $\alpha<\omega_1^{\mathrm{ck}}$).
Then (a) $N\models$ KP$_{\omega_0}$ + $\Sigma_1$-Induction over $\omega$ + "$V=L$", and hence models $\Pi_1$-Induction, but (b) $N\not\models\Sigma_1$-Induction.
To see (a): The main issue is $\Sigma_1$-bounding.
So let $\varphi$ be $\Sigma_1$ and $p,d\in N$,
and suppose that $N\models$"For all $x\in d$, there is $y$ such that $\varphi(p,x,y)$".
Let $\gamma\in\mathrm{OR}^M$
with $\gamma$ illfounded.
Then $M\models$"For all $x\in d$
there is $\alpha<\gamma$
such that $L_{\eta+\alpha}\models\varphi(p,x,y)$",
and so the function $x\mapsto\alpha_x$,
where $\alpha_x$ is the least $M$-ordinal such $\alpha$
such that $L_{\eta+\alpha}^M\models\varphi(p,x,y)$,
is $\Sigma_1^M(M)$,
and hence is an element of $M$.
But $\alpha_x<\omega_1^{\mathrm{ck}}$ for each $x\in M$. Since $\omega_1^{\mathrm{ck}}\notin M$,
this function must be bounded
in $\omega_1^{\mathrm{ck}}$,
which suffices.
Finally, to see (b), just use the fact that "ordinal" addition
is $\Sigma_1$-definable, but
the class of all $\alpha\in\mathrm{OR}^N$ such that $N\models$"$\eta+\alpha$ exists" is just the wellfouned part of $N$, i.e. $\omega_1^{\mathrm{ck}}$,
so $N\models$"for all sets $x$,
if $x$ is an ordinal and for all $y\in x$, $\eta+y$ exists,
then $\eta+x$ exists",
but $N\not\models$"for all sets $x$,
if $x$ is an ordinal then $\eta+x$ exists".
